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:
[tex] E_2 = 2 E_1[/tex]
Explanation:
given,
Charge,q = +0.2 μC
using equation of electric field
[tex]E = \dfrac{kq}{r^2}[/tex]
r is the distance of the position where electric field is to be calculated.
k is coulomb constant
when q = +0.2 μC
[tex]E_1 = \dfrac{k\times +0.2 \muC}{r^2}[/tex]
when charge is changed to q' = +0.4 μC
distance is same r' = r
[tex]E_2= \dfrac{k\times +0.4 \muC}{r^2}[/tex]
[tex]E_2= 2\times \dfrac{k\times +0.2 \muC}{r^2}[/tex]
[tex] E_2 = 2 E_1[/tex]
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.
