Question

Calculate the intensity of electric field due to the spear at the following point at the surface of the sphere at an external point is 0.8 from the surface of the spear at a point of 0.4 metre from the centre of the sphere

Answer 1

To Find:

Intensity of electric field due to the sphere

• At the surface of the sphere.
• At an external point is 0.8m from the surface of the sphere.
• At a point of 0.4 m from the centre of the sphere.

Calculation:

Let's assume that charge on sphere be q and it's Radius be r :

1. At the surface of the sphere:

$\therefore \: E = \dfrac{kq}{ {r}^{2} }$

$\boxed{ = > \: E = \dfrac{q}{4\pi \epsilon_{0}{r}^{2} } }$

2. At 0.8 m from the surface of sphere:

$\therefore \: E = \dfrac{kq}{ {d}^{2} }$

$= > \: E = \dfrac{kq}{ {(r + 0.8)}^{2} }$

$\boxed{ = > \: E = \dfrac{q}{4\pi \epsilon_{0}{(r + 0.8)}^{2} } }$

3. At 0.4 m from centre of sphere:

Considering r > 0.4 m:

$\therefore \: E = \dfrac{kqx}{ {r}^{3} }$

$= > \: E = \dfrac{kq(0.4)}{ {r}^{3} }$

$\boxed{ = > \: E = \dfrac{(0.4)q}{4\pi \epsilon_{0}{r}^{3} } }$

Hope It Helps

Answer 2

AnswEr :

Let the radius of the sphere be R.

To calculate the Electric Field Intensity :

• On the surface
• 0.8 m metres from the sphere at an external point
• 0.4. from the centre of the sphere

Electric Field Intensity in a uniformly charged sphere is given as :

$\large{\star \ \boxed{\boxed{\sf E = \dfrac{1}{4\pi \epsilon_o} \times \dfrac{q}{r^2}}}}$

For 'E' at the surface of the sphere,

r = R

Thus,

• $\sf E = \dfrac{1}{4\pi \epsilon_o} \times \dfrac{q}{R^2}$

For an external point 0.8m away from the surface,

r = (R + 0.8) m

Thus,

• $\sf E = \dfrac{1}{4\pi \epsilon_o} \times \dfrac{q}{(R + 0.8)^2}$

For a point 0.4m away from the centre of sphere,

• $\sf E = \dfrac{1}{4\pi \epsilon_o} \times \dfrac{0.4q}{(R)^2}$