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JEE • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential and capacitance practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 121-150 of 407 questions

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121 Medium
121 of 407
Understanding +4 | -1
Which of the following is not true?
View answer and explanation
Correct option: C
The electric dipole potential varies as 1/r1 / r at large distance
Option C (The electric dipole potential varies as 1/r1 / r at large distance) is correct because dipole potential varies as 1/r21/r^2.
122 Medium
122 of 407
Understanding +4 | -1
A point charge qq is rotated along a circle in the electric field generated by another point charge Q.Q. The work done by the electric field on the rotating charge in one complete revolution is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the electric field is conservative, and the work done by a conservative force in a closed loop is zero.
123 Medium
123 of 407
Understanding +4 | -1
A hollow metal sphere of radius 5 cm5 \mathrm{~cm} is charged such that the potential on its surface is 10 V10 \mathrm{~V}. The potential at the centre of the sphere is
View answer and explanation
Correct option: B
10 V10 \mathrm{~V}
Option B (10 V10 \mathrm{~V}) is correct because the potential inside a hollow charged sphere is constant and equal to the potential on its surface.
124 Medium
124 of 407
Understanding +4 | -1
When an electric dipole p\mathbf{p} is placed in a uniform electric field E\mathbf{E}, then at what angle between p\mathbf{p} and EE the value of torque will be maximum?
View answer and explanation
Correct option: A
90∘90^{\circ}
Option A is correct. The torque on an electric dipole in a uniform electric field is maximum when the angle between the dipole moment and the electric field is 90 degrees, as τ=pEsin⁡θ\tau = pE\sin\theta.
125 Medium
125 of 407
Understanding +4 | -1
If a dielectric substance is introduced between the plates of a charged air-gap capacitor. The energy of the capacitor will
View answer and explanation
Correct option: B
Decrease
Option B (Decrease) is correct because introducing a dielectric increases capacitance, and since the charge is constant, the energy decreases (U=Q22CU = \frac{{Q^2}}{{2C}}).
126 Medium
126 of 407
Understanding +4 | -1
If a positive charge is shifted from a low-potential region to a high-potential region, then energy is
View answer and explanation
Correct option: B
Increases
Option B (Increases) is correct because the potential energy of a positive charge increases when moved from a low to a high potential region.
127 Medium
127 of 407
Analyzing +4 | -1
Four charges +Q,−Q,+Q+Q,-Q,+Q and −Q-Q are situated at the corners of a square; in a sequence then at the centre of the square
View answer and explanation
Correct option: C
E≠0, V=0{E} \neq 0, {~V}=0
Option C (E≠0, V=0{E} \neq 0, {~V}=0) is correct because the electric field at the center due to the charges does not cancel out, but the electric potential does due to the alternating signs of the charges.
128 Medium
128 of 407
Application +4 | -1
Two charged particles, having equal charges of 2.0×10−5C2.0 \times 10^{-5} \mathrm{C} each, are brought from infinity to within a separation of 10 cm10 \mathrm{~cm}. Find the increase in the electric potential energy during the process.
View answer and explanation
Correct option: D
36 J
Option D (36 J) is correct because the increase in electric potential energy is calculated using the formula U=kq1q2rU = k \frac{q_1 q_2}{r}.
129 Easy
129 of 407
Remembering +4 | -1
One micro microfarad is equal to
View answer and explanation
Correct option: B
1 pF1 \,{pF}.
Option B (1 pF1 \,{pF}) is correct because micro microfarad is equivalent to picofarad.
130 Medium
130 of 407
Remembering +4 | -1
The capacitance of an isolated conducting sphere of radius RR is proportional to
View answer and explanation
Correct option: D
RR
Option D (RR) is correct because the capacitance of an isolated conducting sphere is directly proportional to its radius.
131 Medium
131 of 407
Application +4 | -1
Three capacitors of 2μF,  3μF2\mu F,\;3\mu F and 6μF6\mu F are joined in series and the combination is charged by means of a 24  volt24\;volt battery. The potential difference between the plates of the 6μF6\mu F capacitor is
View answer and explanation
Correct option: A
4  volt4\;volt
Option A (4  volt4\;volt) is correct because the potential difference across the 6μF6\mu F capacitor is calculated using the total charge and the equivalent capacitance of the series combination.
132 Medium
132 of 407
Application +4 | -1
Three concentric spherical shells have radii a, b and c(a<b<c)\mathrm{c}(\mathrm{a} < \mathrm{b} < \mathrm{c}) and have surface charge densities σ,−σ\sigma,-\sigma and σ\sigma respectively. If VA,VBV_A, V_B and VC\mathrm{V}_{\mathrm{C}} denote the potentials of the three shells, then, for VA=VCV_A=V_C, we get -
View answer and explanation
Correct option: D
c=a+bc=a+b
Option D (c=a+bc=a+b) is correct because it satisfies the condition VA=VCV_A = V_C when the potentials are calculated for the three concentric shells.
133 Medium
133 of 407
Understanding +4 | -1
Two charges +q and –qq are kept apart. Then at any point on the right bisector of line joining the two charges
View answer and explanation
Correct option: B
The electric potential is zero
Option B is correct because the electric potential is a scalar quantity and the potentials due to +q and -q cancel out at any point on the right bisector.
134 Medium
134 of 407
Application +4 | -1
The displacement of a charge QQ in the electric field E⃗=e1i^+e2j^+e3k^\vec E = {e_1}\hat i + {e_2}\hat j + {e_3}\hat k is r⃗=ai^+bj^\vec r = a\hat i + b\hat j. The work done is
View answer and explanation
Correct option: A
Q(ae1+be2)Q\left( {a{e_1} + b{e_2}} \right)
Option A (Q(ae1+be2)Q\left( {a{e_1} + b{e_2}} \right)) is correct because the work done is the dot product of the force (Q*E) and the displacement vector.
135 Medium
135 of 407
Analyzing +4 | -1
Two spheres of radii rr and RR carry charges qq and QQ respectively. When they are connected by a wire, there will be no loss of energy of the system if
View answer and explanation
Correct option: B
qR=Qr{qR}={Qr}
The correct answer is B (qR=Qr{qR}={Qr}) because for no energy loss, the potentials of the two spheres must be equal after they are connected.
136 Medium
136 of 407
Application +4 | -1
The work done in bringing a 20 coulomb charge from point AA to point BB for distance 0.2 m is 2 joule. The potential difference between the two points will be (in volt)
View answer and explanation
Correct option: C
0.1
Option C (0.1) is correct because the potential difference is calculated as the work done per unit charge, V=WQV = \frac{W}{Q}.
137 Medium
137 of 407
Application +4 | -1
Three charges each 20μC20\mu C are placed at the corners of an equilateral triangle of side 0.4m0.4m. The potential energy of the systems is…… JJ
View answer and explanation
Correct option: A
27
Option A (27) is correct because it is the potential energy calculated using the formula for potential energy of a system of charges.
138 Medium
138 of 407
Application +4 | -1
Two square plates are at potential difference of 100 V separated by 2 cm. calculate electric intensity between them.
View answer and explanation
Correct option: B
5000
Option B (5000) is correct because the electric field intensity is calculated by dividing the potential difference by the distance between the plates.
139 Medium
139 of 407
Understanding +4 | -1
The electric field at a point on the equatorial plane at a distance r\text{r} from the centre of a dipole having dipole moment P\text{P} is given by (r≫(\text{r} \gg separation of two charges forming the dipole, ε0\varepsilon_0= permittivity of free space)
View answer and explanation
Correct option: D
E=−P4πε0r3\mathrm{E}=-\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^3}
Option D (E=−P4πε0r3\mathrm{E}=-\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^3}) is correct because it represents the electric field at a point on the equatorial plane of a dipole.
140 Medium
140 of 407
Application +4 | -1
A capacitor has capacitance 2 F2 \mathrm{~F}. plate separation 0.5 cm0.5 \mathrm{~cm} then area of plate
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated area is 1.130×109m21.130 \times 10^9 m^2, which is not among the given options.
141 Medium
141 of 407
Application +4 | -1
On moving a charge of 30 coulombs by 3 cm, 6 J of work is done, then the potential difference between the points is
View answer and explanation
Correct option: A
0.2 V0.2 \mathrm{~V}
Option A (0.2 V0.2 \mathrm{~V}) is correct because the potential difference is the work done per unit charge, which is calculated as 6 J / 30 C.
142 Medium
142 of 407
Application +4 | -1
1000 small water drops each of radius rr and charge qq coalesce together to form one spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of
View answer and explanation
Correct option: B
100
Option B (100) is correct because the potential of a larger drop formed by coalescing smaller drops is proportional to n(2/3), where n is the number of smaller drops.
143 Medium
143 of 407
Understanding +4 | -1
An insulated charged sphere of radius 5 cm has
a potential of 10 V at the surface. The potential
at centre will be
View answer and explanation
Correct option: C
10 V
Option C is correct because the potential inside a charged sphere is constant and equal to the potential at the surface.
144 Medium
144 of 407
Application +4 | -1
If 20 J20 \mathrm{~J} of work has to be done to move an electric charge of 4C4 \mathrm{C} from a point, where potential is 10 V10 \mathrm{~V} to another point, where potential is V\mathrm{V}, find the value of V\mathrm{V}.
View answer and explanation
Correct option: D
15 V15 \mathrm{~V}
Option D (15 V) is correct because the potential difference multiplied by the charge equals the work done.
145 Medium
145 of 407
Application +4 | -1
A 4  μF4\,\,\mu F conductor is charged to 400  V\mathrm{400}\,\,\mathrm{V} and then its plates are joined through a resistance of 1  KΩ\mathrm{1}\,\,\mathrm{K\Omega }. The heat produced in the resistance is
View answer and explanation
Correct option: D
0.32  J\begin{array}{llllllllllllllllllll} {\mathrm{0}\mathrm{.32}\,\,\mathrm{J}} \end{array}
Option D (0.32 J) is correct because the energy stored in the capacitor is converted into heat in the resistor, and the energy stored is calculated by 12CV2\frac{1}{2}CV^2.
146 Medium
146 of 407
Application +4 | -1
A solid sphere of radius R\mathrm{R} is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre
View answer and explanation
Correct option: C
R3\frac{R}{3}
Option C (R/3R/3) is correct because it is the distance from the surface where the potential is half the potential at the center.
147 Medium
147 of 407
Understanding +4 | -1
The relation between the intensity of the electric field of an electric dipole at a distance r from its centre on its axis and the distance r is where (r≫2l)r \gg 2l)
View answer and explanation
Correct option: D
E∝1r3E \propto \frac{1}{{{r^3}}}
Option D (E∝1r3E \propto \frac{1}{{{r^3}}}) is correct because the electric field due to a dipole at a distance r on its axis is inversely proportional to r cubed when r is much greater than 2l.
148 Medium
148 of 407
Application +4 | -1
If 7 charge qq are placed at corners of cube of edge length 3 m\sqrt{3} \mathrm{~m} then find potential at centre of cube:
View answer and explanation
Correct option: B
7q6πε0\frac{7 q}{6 \pi \varepsilon_0}
Option B (7q6πε0\frac{7 q}{6 \pi \varepsilon_0}) is correct because it is the electric potential at the center of the cube due to the 7 charges placed at its corners, calculated using the formula for electric potential due to a point charge.
149 Medium
149 of 407
Understanding +4 | -1
Which one statement is correct? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are Q0,V0,E0Q_0, V_0, E_0 and U0U_0 respectively. In order to fill the complete space between the plates a dielectric slab is inserted, the battery is still connected. Now the corresponding values Q,V,EQ, V, E and UU are in relation with the initially stated as
View answer and explanation
Correct option: A
Q>Q0Q>Q_0
Option A (Q>Q0Q>Q_0) is correct because when a dielectric slab is inserted with the battery connected, the voltage remains constant, but the capacitance increases, leading to an increase in charge.
150 Medium
150 of 407
Application +4 | -1
The capacitance of a metallic sphere will be 1μF1\mu F, if its radius is nearly
View answer and explanation
Correct option: A
9km9km
Option A (9km9km) is correct because it is calculated using the formula for the capacitance of a sphere, solving for the radius.
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