The drawing shows a positive point charge + q 1 , a second point charge q 2 that
ID: 2283312 • Letter: T
Question
The drawing shows a positive point charge +q1, a second point charge q2that may be positive or negative, and a spot labeled P, all on the same straight line. The distance d between the two charges is the same as the distance between q1 and the spot P. With q2 present, the magnitude of the net electric field at P is twice what it is when q1 is present alone. Given that q1 = +3.25 The drawing shows a positive point charge +q1, a second point charge q2that may be positive or negative, and a spot labeled P, all on the same straight line. The distance d between the two charges is the same as the distance between q1 and the spot P. With q2 present, the magnitude of the net electric field at P is twice what it is when q1 is present alone. Given that q1 = +3.25Explanation / Answer
q1 = 3.25uC = 3.25 x 10^-6 C
E = Kq/r^2
E due to q1 = kq1/d^2
= 9 x 3.25 x 10^(9-6)/d^2
= 29.25 x 10^3/d^2
E due to q2 = kq2/(2d)^2
= 9 x 10^9 x q2/4d^2
= 2.25 x 10^9 x q2/d^2
if q2 is positive then
Enet = E due to q1 + E due to q2
and also Enet = 2 E due to q1
Edue to q1 = E due to q2
29.25 x 10^3/d^2 = 2.25 x 10^9 x q2 /d^2
q2 = 29.25 x 10^(3-9)/2.25
q2 = 13 x 10^-6 C
q2 = 13 uC
if q2 is negative then
Enet = Edue to q2 - E due to q1
Enet = 2 E due to q1
E due to q2 = 3E due to q1
2.25 x 10^9 x q2 /d^2 = 3 x 29.25 x 10^3/d^2
q2 = 87.75 x 10^(3-9)/2.25
q2 = 39 x 10^-6 C
q2 = 39 uC
since q2 i s negative
q2 = -39 uC
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.