Question

A hollow metallic sphere of 8cm diameter is charged with 4×10^-8 the potential on its surface will be?

Answer 1

Answer:

9000

Explanation:

v=q/c

q/4 pie ε0 r

(9 exponent 9 × 4 exponent -8) / (4 exponent-2)

9000

Answer 2

Concept:

The potential on surface of a sphere with charge q and radius r is

$V=\frac{Q}{4\pi\varepsilon_0R}$

Given:

Q=4×10^-8C

∵ Diameter = 8cm

∴ R=4cm =0.04 m

Find:

Electric potential(V) on the surface of sphere

Solution:

$V=\frac{Q}{4\pi\varepsilon_0R}$

We know that the value of Coulomb’s constant is:

$\frac{1}{4\pi\varepsilon_0}=9\times{10}^9Nm$

$V=\frac{9\ast{10}^9\ast\left(-4\right)\ast{10}^{-8}}{4\times{10}^{-2}}$

$V=-9000V$

The electric potential on the surface on the hollow cylinder with radius 4 cm is -9000V

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