Question

To what potential we must charge insulated spher of redius 14cm so that surface charge density is equal to 1 u

Answer 1

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QUESTION :

To what potential we must charge insulated spher of redius 14cm so that surface charge density is equal to 1 u...

SOLUTION :

We know that the following Formulas :

Electric Potential , V is equal to k Q / r.

Q = $\sigma$ 4 π r ^ 2

K is equal to 1 / { 4 π e 0 }

K is equal to 1 / { 4 π e 0 }This is equal to 9 × 10 ^ - 9

K is equal to 1 / { 4 π e 0 }This is equal to 9 × 10 ^ - 9Now, Substituting the required values :

V = 9 × 10 ^ -9 × 1 × π × 14 ^ 2 / 14

=> V = 9 π × 14 × 10 ^ -9

=> V = 136 π × 10 ^ -9

ANSWER :

V = 136 π × 10 ^ -9.

Answer 2

The electric potential is 15825.6 volts.

Given,

An insulated sphere of radius 14 cm

To Find,

To what potential we must charge the sphere so that the surface charge density is equal to 1 u?

Solution,

We can solve the question as follows:
It is given that the radius of the sphere is 14 cm. The surface charge density is 1 u.

$Radius = 14\: cm = 0.14\: m$

$Surface\: charge\: density,$ σ $= 1$ μ

First, we will find the charge present in the sphere.

The formula for calculating the surface charge density is:

σ $= \frac{q}{A}$

Where, $A = Surface\: area\: of\: the\: sphere$

Substituting the values,

σ $= \frac{q}{4\pi r^{2} }$

$10^{-6} = \frac{q}{4\pi (0.14)^{2} }$

$q = 10^{-6} *4\pi (0.14)^{2}$

Now, the formula for finding the electric potential is:

$V = \frac{kq}{r}$

Where, $k = 9 * 10^{9}$

Substituting the values,

$V = \frac{9*10^{9} *10^{-6}*4*\pi *0.14^{2} }{0.14}$

$V = 36*3.14*10^{3} *0.14$

$V = 15825.6\: volts$

Hence, the electric potential is 15825.6 volts.

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