Question

If 10c charge is placed on the surface of a sphere having 9x10''V electric potential at its
surface, what is the radius of the sphere?​

Answer 1

Given :

• Charge placed on the surface of sphere , Q= 10 C
• Electric Potential at its surface$\sf=9\times10^{11}V$

To Find :

The Radius of the sphere

Theory :

• Electric potential due to uniformly charged shell / solid conductor or sphere

$\sf\:1)V_{in}=\dfrac{KQ}{R}$ , r< R

$\sf\:2)V_{surface}=\dfrac{KQ}{R}$ , r = R

$\sf\:3)V_{out}=\dfrac{KQ}{r}$ , r > R

Where , R = Radius of the sphere or shell

• Electric potential due to uniformly non- conducted sphere

$\sf\:1)V_{in}=\dfrac{KQ}{2R^3}(3R^2-r^2)$

$\sf\:2)V_{surface}=\dfrac{KQ}{R}$

$\sf\:3)V_{centre}=\dfrac{3}{2}\times\dfrac{KQ}{r}$

Solution :

We have to Find thr Raduis of the sphere

We know that

Electric Potential due to uniformly charged sphere :

$\sf\blue{V=\dfrac{KQ}{R}}$

$\sf\implies\:R=\dfrac{KQ}{V}$

Now , Put the given values

$\sf\implies\:R=\dfrac{9\times10^{9}\times10}{9\times10^{11}}$

$\sf\implies\:R=\dfrac{9\times10^{10}}{9\times10^{11}}$

$\sf\implies\:R=\dfrac{1}{10}$

$\sf\implies\:R=0.1m$

Therefore, the Raduis of the given sphere is 0.1m

Answer 2

Answer:

Given :

Charge placed on the surface of sphere , Q= 10 C

Electric Potential at its surface\sf=9\times10^{11}V=9×10

11

V

To Find :

The Radius of the sphere

Theory :

• Electric potential due to uniformly charged shell / solid conductor or sphere

\sf\:1)V_{in}=\dfrac{KQ}{R}1)V

in

=

R

KQ

, r< R

\sf\:2)V_{surface}=\dfrac{KQ}{R}2)V

surface

=

R

KQ

, r = R

\sf\:3)V_{out}=\dfrac{KQ}{r}3)V

out

=

r

KQ

, r > R

Where , R = Radius of the sphere or shell

• Electric potential due to uniformly non- conducted sphere

\sf\:1)V_{in}=\dfrac{KQ}{2R^3}(3R^2-r^2)1)V

in

=

2R

3

KQ

(3R

2

−r

2

)

\sf\:2)V_{surface}=\dfrac{KQ}{R}2)V

surface

=

R

KQ

\sf\:3)V_{centre}=\dfrac{3}{2}\times\dfrac{KQ}{r}3)V

centre

=

2

3

×

r

KQ

Solution :

We have to Find thr Raduis of the sphere

We know that

Electric Potential due to uniformly charged sphere :

\sf\blue{V=\dfrac{KQ}{R}}V=

R

KQ

\sf\implies\:R=\dfrac{KQ}{V}⟹R=

V

KQ

Now , Put the given values

\sf\implies\:R=\dfrac{9\times10^{9}\times10}{9\times10^{11}}⟹R=

9×10

11

9×10

9

×10

\sf\implies\:R=\dfrac{9\times10^{10}}{9\times10^{11}}⟹R=

9×10

11

9×10

10

\sf\implies\:R=\dfrac{1}{10}⟹R=

10

1

\sf\implies\:R=0.1m⟹R=0.1m

Therefore, the Raduis of the given sphere is 0.1m

Explanation:

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