Question

The magnitude of electric field at a distance x from a charge q is e.An identical charge is placed at a distance 2x from it. Find the magnitude of force it experiences

Answer 1

Electric field at distance 2x will E/4.

Explanation is given in image

Answer 2

Answer:

The magnitude of force experience by charge, q at a distance 2x is qe/4 .

Explanation:

• The magnitude of electric field due to charge, q at a distance, r is

$E = k \frac{q}{ {r}^{2} }$

The force experience by a charge q2 at a distance r from charge q1 is given by:

$F = k \frac{q1q2}{ {r}^{2} }$

Given that :

• Electric field at distance x due to charge q is e.

To find :

• Magnitude of force on charge, q at distance 2x

Solution:

• The magnitude of force experience by charge, q at a distance 2x from another charge, q is

$F = k \frac{ {q}^{2} }{ {(2x)}^{2} } \\ = \frac{k}{4} \frac{ {q}^{2} }{ {x}^{2} } \: - - - (1)$

• Also, Electric field due to charge, q at point x is e.

$e = k \frac{q}{ {x}^{2} } \: - - - (2)$

• From equation (1) and (2),

$F = \frac{k}{4} \frac{ {q}^{2} }{ {x}^{2} } = \frac{qe}{4}$

• Hence, the magnitude of force is qe/4.