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
Answer:
q1 = 7.25×10-¹⁶C
Explanation:
Fd = Drag force = 7.25×10-¹¹N, E = 1.00×10⁵N/C
The particle moves with constant velocity so by Newtown's first law of motion the sum of forces acting of the charge is zero. Two forces act on the charge
Fe – Fd = 0
Fd is negative because it is against the direction of motion
Fe = q1×E
So
q1×E – Fd = 0
q1×E = Fd
q1 = Fd/E = 7.25×10-¹¹/(1.00×10⁵)
q1 = 7.25×10-¹⁶C
