Two point charges are placed on the x-axis. The first charge,q1 = 8.00 nC, is pl

Question

Two point charges are placed on the x-axis. The first charge,q1 = 8.00 nC, is placed a distance 16.0 m from the origin along the positive x-axis; thesecond charge, q2 = 6.00 nC, is placed a distance 9.00 m from the origin along the negativex-axis. Find the magnitude and sign of q3 needed to make thetotal electric field at point A equal to zero. Two point charges are placed on the x-axis. The first charge,q1 = 8.00 nC, is placed a distance 16.0 m from the origin along the positive x-axis; thesecond charge, q2 = 6.00 nC, is placed a distance 9.00 m from the origin along the negativex-axis. Find the magnitude and sign of q3 needed to make thetotal electric field at point A equal to zero.

Explanation / Answer

Let the point A be at a distance x from the chargeq1 along the positive x-axis The electric field due to the charge q1 is E1 = (1/4o) *(q1/x2) The electric field due to the charge q2 is E2 = (1/4o) *(q2/(25 + x)2) The total electric field at point A due to the chargesq1 and q2 is zero therefore we get E1 + E2 = 0 or (1/4o) *(q1/x2) + (1/4o) *(q2/(25 + x)2) = 0 or (q1/x2) = -(q2/(25 +x)2) or -(25 + x/x)2 =(q2/q1) or -(25/x + 1) = (q2/q1)1/2 or (25/x + 1) =-(q2/q1)1/2 or (25/x) = -((q2/q1)1/2 +1) or x = -(25/((q2/q1)1/2 +1)) where q2 = 6.00 nC = 6.00 * 10-9 C andq1 = 8.00 nC = 8.00 * 10-9 C The negative value of distance indicates that the chargeq3 should be placed in the opposite direction to thedirection we have considered. or (25/x + 1) =-(q2/q1)1/2 or (25/x) = -((q2/q1)1/2 +1) or x = -(25/((q2/q1)1/2 +1)) where q2 = 6.00 nC = 6.00 * 10-9 C andq1 = 8.00 nC = 8.00 * 10-9 C The negative value of distance indicates that the chargeq3 should be placed in the opposite direction to thedirection we have considered.

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