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Question

Two point charges Q1 and Q2 are positioned at points 1 and 2, The field intensity to the right of the charge Q2 on the line that passes through the two charges varies according to a law that is represented schematically in the figure.

The field intensity is assumed to be positive if its direction coincides with the positive direction on the x−axis.The distance between the charges is l. Find the sign of each charge.

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Solution

The correct option is **D** Q2 is negative and Q1 is positive

Since, field intensity at point 2 that is at location of Q2 is −∞.

E=Q4πϵor

Hence, Q2 must be negative and only then for r=0, field at 2 will be −∞.

Now at some distance right of point 2, field is 0.

At that point E=Q14πϵo(l+a)+Q24πϵoa=0

Since, Q2 is negative and hence Q1 must be positive for E to be 0..

Answer-(D)

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