Your answer is C.
hope this helps!
Answer:
Q1 = 7.25*10^(-16) C
Explanation:
We are given;
electric field strength = (1 x 10^5 N/C
drag force (F) = 7.25 x 10^(-11) N
The question says it's moving with constant velocity. This means that he particle is in equilibrium and not accelerating.
Columbs law force of attraction or repulsion between two charges is given as;
F=(KQ1Q2)/r²
Now, electric field strength is given as the formula;(K*Q2)/r², thus plugging the relevant values gives us;
7.25 x 10^(-11) N= (1 x 10^(5) N/C)Q1 Q1 = 7.25 x 10^(-11) /(1 x 10^(5))
Q1 = 7.25*10^(-16) C
No work is done by THAT force.