Question

Three point charges are arranged in a line. Charge q3+5.00 nC and is located at the origin Charge q,--3.00 nC and is located at' +4.00 cm. Charge q! is located at x +2.00 cm What is the magnitude of qı if the net force on q3 1s zero? point) Oo.750 nC O0.520 nC O0.975 nC O0.000 nC 4 Two identical point charges in outer space are held apart at a distance D. As soon as the charges are released, each begins moving at an acceleration a. If instead they were released at a distance _, the acceleration of one charge would be (1 point) 2 O2a 2 4 5. Two unequal charges repel each other with a force of magnitude F. After both charges are tripled in magnitude, the new repelling force between them has magnitude (1 point) 6F 09F O27F

Answer 1

3.

Electrostatic force is given by:

F = k*Q1*Q2/R^2

Where force will be attractive if both charge have opposite signs and force will be repulsive if both charges have same sign

Now net force on charge q3 is given that zero, So

Fnet = F13 - F23 = 0

F13 = Force on q3 due to q1

F23 = Force on q3 due to q2

F13 = F23

k*q3*q1/r1^2 = k*q3*q2/r2^2

q1/q2 = (r1/r2)^2

q1 = q2*(r1/r2)^2

r1 = distance b/w q1 and q3 = 2.00 - 0.00 = 2.00 cm

r2 = distance b/w q2 and q3 = 4 - 0 = 4.00 cm

q2 = -3.00 nC

Magnitude of q1 will be

|q1| = |-3*(2/4)^2|

|q1| = 3/4 = 0.75 nC

Correct option is A.

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4.

From above equation of force

F = k*q1*q2/r^2

From newton's 2nd law: F = m*a

a = k*q1*q2/(r^2*m)

from above equation we can see that acceleration is inversely proportional to the distance between both charges, So

a2/a1 = (r1/r2)^2

a2 = a1*(r1/r2)^2

given that r1 = D & r2 = D/2

and a1 = a, So

a2 = a*(D/(D/2))^2 = a*(2^2)

a2 = 4*a

Correct option is B.

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5.

Electrostatic force is given by:

F = k*q1*q2/r^2

from above equation we can see that force is directly proportional to the magnitude of both the charge, So

F2/F1 = (q1'*q2'/(q1*q2))^2

q1' = 3*q1

q2' = 3*q2

F1 = F

So,

F2/F = (3*q1*3*q2/(q1*q2))

F2/F = 9

F2 = 9*F

Correct option is C.

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