Question

EXIULUI
A hollow charged metal sphere has radius r. If the
potential difference between its surface and a point
at a distance 3r from the centre is V, then the electric
field intensity at distance 3r from the centre is
​

Answer 1

Explanation:

The charge Q on a hollow metal sphere is uniformly distributed on its surface. This means that the potential and electric field outside the sphere are the same as these quantities for a point charge Q placed at the center of the sphere.
Thus, the potential difference between the surface and the point at a distance 3r is
$v = \frac{1}{4\pi \: e _{0} } ( \frac{q}{r} - \frac{q}{3r} )$
$\frac{1}{4\pi \: e _{0} } ( \frac{q}{r} - \frac{q}{3r} ) \: = \frac{2}{3} \frac{1}{4\pi \: e_{0} } \frac{q}{r}$

The electric field at this point is
$E \: = \frac{1}{4\pi \: e _{0} } \frac{ Q \: }{ {3r}^{2} }$
$\frac{1}{4\pi \: e _{0} } \frac{ Q \: }{ {3r}^{2} } \: = \frac{1}{9} \frac{1}{4\pi \: e_{0} } \frac{ Q \: }{ {r}^{2} }$
$\frac{1}{9} \frac{1}{4\pi \: e_{0} } \frac{ Q \: }{ {r}^{2} } \: = \frac{1}{9} \frac{ \frac{3}{2} v}{r}$
$\frac{1}{9} \frac{ \frac{3}{2} v}{r} = \frac{1}{6} \frac{v}{r}$
hope it helps you

Answer 2

Answer:

The charge Q on a hollow metal sphere is uniformly distributed on its surface. This means that the potential and electric field outside the sphere are the same as these quantities for a point charge Q placed at the center of the sphere.

Thus, the potential difference between the surface and the point at a distance 3r is

v = \frac{1}{4\pi \: e _{0} } ( \frac{q}{r} - \frac{q}{3r} )v=

4πe

0

1

(

r

q

−

3r

q

)

\frac{1}{4\pi \: e _{0} } ( \frac{q}{r} - \frac{q}{3r} ) \: = \frac{2}{3} \frac{1}{4\pi \: e_{0} } \frac{q}{r}

4πe

0

1

(

r

q

−

3r

q

)=

3

2

4πe

0

1

r

q

The electric field at this point is

E \: = \frac{1}{4\pi \: e _{0} } \frac{ Q \: }{ {3r}^{2} }E=

4πe

0

1

3r

2

Q

\frac{1}{4\pi \: e _{0} } \frac{ Q \: }{ {3r}^{2} } \: = \frac{1}{9} \frac{1}{4\pi \: e_{0} } \frac{ Q \: }{ {r}^{2} }

4πe

0

1

3r

2

Q

=

9

1

4πe

0

1

r

2

Q

\frac{1}{9} \frac{1}{4\pi \: e_{0} } \frac{ Q \: }{ {r}^{2} } \: = \frac{1}{9} \frac{ \frac{3}{2} v}{r}

9

1

4πe

0

1

r

2

Q

=

9

1

r

2

3

v

\frac{1}{9} \frac{ \frac{3}{2} v}{r} = \frac{1}{6} \frac{v}{r}

9

1

r

2

3

v

=

6

1

r

v