Back to ELECTRIC FIELDS AND CHARGES

JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

1 Medium

1 of 1

Application +4 | -1

A tin nucleus (atomic number Z = 50) has a radius of

6.6 \times {10^{ - 15}}\,m

. The potential on its surface will be (the charge on the proton

= 1.6 \times {10^{ - 19}}\,C

)

View answer and explanation

Correct option: A

1.1 \times {10^7}\,\mathrm{V}

Option A (

1.1 \times {10^7}\,\mathrm{V}

) is correct because it accurately calculates the electric potential on the surface of the tin nucleus using the formula

V = \frac{{1}}{{4\pi \varepsilon _0}} \cdot \frac{q}{r}

.

More Questions from ELECTRIC FIELDS AND CHARGES

Practice All Chapter MCQs →

Assertion:- Electric field lines do not cut at any point.
Reason:- At any point, there is only one direction of electric field. Statement-I: A positive point charge is brought in an electric field, the electric field at neary by point will increase

Statement-II: Electric field produced by charge may favour the existing electric filed Two particles having equal mass and charge
are separated by a distance

r.

The force(s) they
experience is (are) Statement-I: No 2 electric lines of force can intersect each other

Statement-II: tangent at any point of electric line of force gives the direction of electric field A force of

4N

is acting between two charges in air if the space between them is completely filled with glass

\left( {{\varepsilon _r} = 8} \right)

, then the new force will be ……

N

An electron is placed at centre of a conducting sphere of radius 0.2 metre having charge

5 \times {10^{ - 2}}

coulomb. The force of on the electron is ________ N