Question

A +0.2 μC charge is at some point P in an electric field. Part A What happens if that charge is replaced by a +0.4 μC charge?

Answer 1

Explanation:

Since, the magnitude of charge becomes twice from 0.2 μC to 0.4 μC, the magnitude of force experienced will also double as

Force F =EQ

F∝Q

Since, Q is doubled , Force experienced by charge will also double.

This force will be force of repulsion since the charges are positive.

Answer 2

Answer:

$E_2 = 2 E_1$

Explanation:

given,

Charge,q = +0.2 μC

using equation of electric field

 $E = \dfrac{kq}{r^2}$

r is the distance of the position where electric field is to be calculated.

k is coulomb constant

when q =  +0.2 μC

 $E_1 = \dfrac{k\times +0.2 \muC}{r^2}$

when charge is changed to q' = +0.4 μC

         distance is same r' = r

 $E_2= \dfrac{k\times +0.4 \muC}{r^2}$

 $E_2= 2\times \dfrac{k\times +0.2 \muC}{r^2}$

$E_2 = 2 E_1$

From, the above equation we can conclude that when the charge +0.2 μC is replaced by +0.4 μC the electric field of the system is doubled.