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Question

The variation of electric field between two charges q1 and q2 along the line joining the charges is plotted against the distance r from q1 (taking the direction from q1 to q2 as positive) as shown in the figure. Find the correct statement among the following is

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Solution

The correct option is **C** q1 is positive and q2 is negative |q1|<|q2|

Just to the right of q1, →E is positive, so we can say that q1 is positive.

Just to the left of q2, →E is negative, so we can say that q2 is negative.

Let the distance between charges q1 and q2 be d and let x be the distance from q1 where net electric field is equal to zero.

∴kq1x2=kq2(d−x)2

⇒∣∣∣q2q1∣∣∣=(d−x)2x2

Further, as →Enet=0 is close to q1, we can conclude that the magnitude of charge q2 must be greater than the magnitude of charge q1.

So, |q2|>|q1|

Hence, option (c) is the correct answer.

Just to the right of q1, →E is positive, so we can say that q1 is positive.

Just to the left of q2, →E is negative, so we can say that q2 is negative.

Let the distance between charges q1 and q2 be d and let x be the distance from q1 where net electric field is equal to zero.

∴kq1x2=kq2(d−x)2

⇒∣∣∣q2q1∣∣∣=(d−x)2x2

Further, as →Enet=0 is close to q1, we can conclude that the magnitude of charge q2 must be greater than the magnitude of charge q1.

So, |q2|>|q1|

Hence, option (c) is the correct answer.

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