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ExamsJEE AdvancedPhysics

A positive charge +q1 is placed to the left of a negative charge -q2 (magnitude q2 = 12 micro-C). On the line joining the two charges, the net electric potential is zero at exactly two points. The first zero-potential point is between the two charges, 4.00 cm to the left of the negative charge. The second zero-potential point is 7.00 cm to the right of the negative charge. Find the value of q1 in micro-C.

  1. 44 micro-C
  2. 36 micro-C
  3. 22 micro-C
  4. 48 micro-C

Correct answer: 44 micro-C

Solution

Let d = separation between +q1 and -q2. Point 1 (between charges, 4 cm from q2): distance from q1 is (d-4). Potential condition: q1/(d-4) = q2/4. Point 2 (7 cm right of q2): distance from q1 is (d+7). Potential condition: q1/(d+7) = q2/7. Dividing: (d+7)/(d-4) = 7/4 gives 4d + 28 = 7d - 28, so d = 56/3 cm. Then q1 = q2*(d-4)/4 = 12*(56/3 - 4)/4 = 12*(44/3)/4 = 44 micro-C.

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