Back to ELECTRIC POTENTIAL AND CAPACITANCE
Your Progress
0 / 372 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE
Electric potential and capacitance practice · Medium
Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.
Practice format and chapter quiz
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.
Take a chapter quiz →
Showing 241-270 of 372 questions
Preparing practice and saved progress…
241
Medium
241 of 372
Application
+4 | -1
A capacitor of unknown capacitance is connected across a battery of volt. The charge stored in it becomes coulomb. When potential across the capacitor is reduced by volt, the charge stored in it becomes coulomb. The capacitance is
View answer and explanation
Correct option: A
Option A () is correct because the capacitance is calculated using the change in charge and the change in voltage across the capacitor.
242
Medium
242 of 372
Understanding
+4 | -1
If combining two charged bodies, the current does not flow then
View answer and explanation
Correct option: C
Potential is equal on both
Option B (Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I) is correct because both statements are true, but they are independent of each other.
243
Medium
243 of 372
Application
+4 | -1
The plates in a parallel plate capacitor are separated by a distance with air as the medium in between the plates. In order to increase the capacity by 66 %, a dielectric slab of dielectric constant 5 is introduced in between the plates. The thick ness of the dielectric slab required is
View answer and explanation
Correct option: B
Option B () is correct because the thickness of the dielectric slab is calculated by equating the new capacitance to 1.66 times the original capacitance.
244
Medium
244 of 372
Application
+4 | -1
Consider the fusion in helium plasma. Find the temperature at which the average thermal energy equals the Coulomb potential energy at .
View answer and explanation
Correct option: A
Option A () is correct because it is the temperature at which the average thermal energy equals the Coulomb potential energy at 2 fm, calculated using the given formula.
245
Medium
245 of 372
Application
+4 | -1
The kinetic energy of a charged particle decreases by as it moves from a point at potential to a point at potential . Find the charge on the particle.
View answer and explanation
Correct option: C
0.1 C
Option C (0.1 C) is correct because the charge is calculated by equating the change in kinetic energy to the work done by the electric potential difference.
246
Medium
246 of 372
Remembering
+4 | -1
If is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to
View answer and explanation
Correct option: B
Option B () is correct because the electrostatic energy density is directly proportional to the square of the electric field intensity.
247
Medium
247 of 372
Understanding
+4 | -1
When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then
View answer and explanation
Correct option: A
Charge on capacitor increases
Option A is correct because introducing a dielectric increases the capacitance, and since the voltage is constant (connected to a battery), the charge on the capacitor increases.
248
Medium
248 of 372
Application
+4 | -1
The electric potential is at a distance of from a point charge Then, is equal to
View answer and explanation
Correct option: A
Option A () is correct because it is the charge calculated using the formula for electric potential due to a point charge.
249
Medium
249 of 372
Application
+4 | -1
Two spheres of radii 12 cm and 16 cm have equal charge. The ratio of their energies is
View answer and explanation
Correct option: A
Option A () is correct because the energy of a charged sphere is inversely proportional to its radius when the charge is constant.
250
Medium
250 of 372
Analyzing
+4 | -1
Assertion (A) : A small electric dipole is moved translationally from higher potential to lower potential in uniform electric field. Work done by electric field is positive.
Reason (R) : When a positive charge is moved from higher potential to lower potential, work done by electric field is positive.
Reason (R) : When a positive charge is moved from higher potential to lower potential, work done by electric field is positive.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D (If Assertion is false but the Reason is true) is correct because the work done by the electric field on a dipole depends on the change in potential energy, and the reason correctly states the work done on a positive charge.
251
Medium
251 of 372
Application
+4 | -1
A sphere of radius 0.03 mis suspended within a hollow sphere of radius . If the inner sphere is charged to a potential of 1500 volt and outer sphere is earthed, find the capacitance and the charge on the inner sphere.
View answer and explanation
Correct option: A
The correct answer is A () because it is the charge calculated using the capacitance of the spherical capacitor and the given potential.
252
Medium
252 of 372
Application
+4 | -1
An electric dipole consists of two opposite charges each 0.05 separated by 30 mm. The dipole is placed in an uniform external electric field of The maximum torque exerted by the field on the dipole is
View answer and explanation
Correct option: D
Option D () is correct because it is calculated using the formula for maximum torque on a dipole in an electric field, , where is the dipole moment and is the electric field.
253
Medium
253 of 372
Application
+4 | -1
A capacitor is charged to 200 V. If its plates are joined through a resistance of then heat produced in the resistance is
View answer and explanation
Correct option: A
0.12 J
Option A (0.12 J) is correct because the heat produced is equal to the energy stored in the capacitor, which is calculated using .
254
Medium
254 of 372
Application
+4 | -1
Taking the Earth to be a spherical conductor of diameter Its capacity will be
View answer and explanation
Correct option: A
Option A () is correct because it is the capacitance calculated using the formula for the capacitance of a spherical conductor.
255
Medium
255 of 372
Application
+4 | -1
If a slab of insulating material thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by to restore the capacity to original value. The dielectric constant of the material will be
View answer and explanation
Correct option: B
8
Option B (8) is correct because the dielectric constant is calculated using the formula relating the thickness of the slab and the increase in separation.
Explore Other Chapters in PHYSICS
View All →
256
Medium
256 of 372
Application
+4 | -1
Electric field is given by potential difference between and .
View answer and explanation
Correct option: A
volt
Option A (5 volt) is correct because the potential difference is calculated by integrating the electric field over the given distance.
257
Medium
257 of 372
Application
+4 | -1
When charge of magnitude is shifted from infinity to a point in an electric field, then it was found that work done by electrostatic forces is . If this charge is doubled and moved again from infinity to the same point without changing its velocity, then find out the work done by electric field and against this electric field.
View answer and explanation
Correct option: A
,
Option A ( ,) is correct because the work done is directly proportional to the charge moved in the electric field.
258
Medium
258 of 372
Understanding
+4 | -1
The electric potential at the centre of a charged
conductor is
conductor is
View answer and explanation
Correct option: D
Same as on the surface
Option D (Same as on the surface) is correct because the electric field inside a charged conductor is zero, implying a constant potential.
259
Medium
259 of 372
Application
+4 | -1
Electric potential at any point is then the magnitude of the electric field is
View answer and explanation
Correct option: D
7
Option D (7) is correct because the electric field components are the negative gradients of the potential, and the magnitude is the square root of the sum of the squares of the components.
260
Medium
260 of 372
Application
+4 | -1
A condenser has a capacity and is charged to a voltage of The energy stored is
View answer and explanation
Correct option: D
Option D () is correct because the energy stored in a capacitor is calculated using the formula E = (1/2)CV², and then converted from Joules to ergs.
261
Medium
261 of 372
Understanding
+4 | -1
A metallic hollow sphere of radius is charged so that potential at its surface is The potential at the centre of sphere will be
View answer and explanation
Correct option: B
Option B () is correct because the electric potential inside a hollow metallic sphere is constant and equal to the potential at the surface.
262
Medium
262 of 372
Understanding
+4 | -1
A ring with uniform distribution of charge and radius is placed in plane with its centre at origin.
View answer and explanation
Correct option: B
Potential at origin is
Option B (Potential at origin is ) is correct because the electric potential at the center of a uniformly charged ring is given by this formula.
263
Medium
263 of 372
Understanding
+4 | -1
The separation between the plates of a charged parallel- plate capacitor is increased. The force between the plates
View answer and explanation
Correct option: C
remains same
Option C (remains same) is correct because the force between the plates depends on the charge density and electric field, which remain constant.
264
Medium
264 of 372
Application
+4 | -1
Capacitance of a parallel plate capacitor becomes times its original value, if a dielectric slab of thickness is inserted between the plates [ is the separation between the plates]. The dielectric constant of the slab is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the dielectric constant is calculated using the formula for capacitance with a dielectric slab inserted.
265
Medium
265 of 372
Application
+4 | -1
Two equal charges ‘q’ of opposite sign are separated by small distance ‘2a’. The electric intensity ‘E’ at a point on the perpendicular bisector of the line joining the two charges at a very distance ‘r’ from the line is
View answer and explanation
Correct option: B
Option B is correct because the electric field due to a dipole at a point on the perpendicular bisector is inversely proportional to the cube of the distance.
266
Medium
266 of 372
Analyzing
+4 | -1
Assertion:- A parallel plate capacitor is connected across a battery through a key. A dielectric slab of dielectric constant is introduced between the plates. The energy which is stored becomes times.
Reason:- The potential difference between the plates remains constant or unchanged.
Reason:- The potential difference between the plates remains constant or unchanged.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because inserting a dielectric increases the stored energy by a factor of K when the capacitor is connected to a battery, and the potential difference remains constant.
267
Medium
267 of 372
Application
+4 | -1
A sample of gas is placed in an electric field of . The dipole moment of each molecule is . Find the maximum torque that can act on a molecule.
View answer and explanation
Correct option: A
Option A () is correct because the maximum torque is calculated by multiplying the dipole moment and the electric field strength.
268
Medium
268 of 372
Application
+4 | -1
You are given an arrangement of three point charges and separated by equal finite distances that electric potential energy of the system is zero. Then the value of is:
View answer and explanation
Correct option: A
Option A (-2/3) is correct because it is the value of x that makes the total electric potential energy of the system zero.
269
Medium
269 of 372
Application
+4 | -1
A uniform electric field of exists in plane making an angle with direction. Consider points and then is:
View answer and explanation
Correct option: A
100 volt
Option A (100 volt) is correct because the potential difference is calculated using the formula .
270
Medium
270 of 372
Application
+4 | -1
The electric intensity due to a dipole of length 10 cm and having a charge of at a point on the axis at a distance 20 cm from one of the charges in air, is
View answer and explanation
Correct option: A
The correct answer is A () because the electric intensity due to a dipole is calculated using the formula .
Revision notes and key takeaways
Review ELECTRIC POTENTIAL AND CAPACITANCE in your textbook or class notes, then use this set to check your understanding.
Showing 241-270 of 372 questions
