Hi. Please be as detailed as possible as this is my practice exam and I need ste

Question

Hi. Please be as detailed as possible as this is my practice exam and I need step by step help to better prepare for the real thing. Thank you SOMUCH!

Situation: Two charges are arrayed to the left of point P at the distances listed. q1 has a positive charge. 0,50m Irom 1.) en q2 is positive, the net electric field at P is twice that of q1 alone. lf q1 is 0.50 x 10r0C what is q2? a. 0.50 x 10-6C b. 1.0 x 10-6C c. 2.0 x 10-6C d. none of these 2.) When q2 is negative, the net electric field at P is twice that of q1 alone. If q1 is 0.50 x 10 6C, what is the magnitude of q2? a. 1.50 x 10-6C b. 6.0 x 10-6C C. 2.5 x 106C d. none of these

Explanation / Answer

1)

E1 = k*q1/r1^2

E2 = k*q2/r2^2

Enet = E1 + E2 = k*(q1/r1^2 + q2/r2^2)

given Enet = 2*E1

2*k*q1/r1^2 = k*(q1/r1^2 + q2/r2^2)

2*q1/r1^2 = (q1/r1^2 + q2/r2^2)

2*0.5*10^-6/0.5^2 = (0.5*10^-6/0.5^2 + q2/1^2)

q2 = 2*10^-6 C

option (d)

=======================

E1x = k*q1/r1^2

E2x = -k*q2/r2^2

Enet = E1x + E2x = k*(q1/r1^2 - q2/r2^2)

given Enet = +/- 2*E1x

2*k*q1/r1^2 = k*(q1/r1^2 - q2/r2^2)

2*q1/r1^2 = (q1/r1^2 - q2/r2^2)

-2*0.5*10^-6/0.5^2 = ((0.5*10^-6/0.5^2) - q2/1^2)

q2 = 2*10^-6 C

-2*0.5*10^-6/0.5^2 = ((0.5*10^-6/0.5^2) - q2/1^2)

q2 = 6*10^-6 N/C

option (b) <<<======answer

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