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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 distance from q1 (taking rightward direction of electric field as positive) as shown in the figure. Then, the correct statement is

A
q1 and q2 are positive and q1<q2
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B
q1 and q2 are positive and q1>q2
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C
q1 and q2 are negative and |q1|<|q2|
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D
q1 is positive and q2 is negative ; q1<|q2|
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Solution

The correct option is A q1 and q2 are positive and q1<q2

We can see from the graph that:

(1).
Just to the right of q1, the electric field is +α (let) or in positive direction (away from q1).
Hence, q1 is positive because r and E both are positive.

(2).
Just to the left of q2, the electric field is α (let) or toward left (away from q2).
Hence, q2 is also positive because r and E both are negative.

(3).
E=0 is near q1.
Hence q1<q2 because Eq and E1r.

Hence, option (a) is the correct answer.


Why this question?
Concept - For positive charge in Ex curve, x positive, E positive and x negative, E negative.
For negative charge in Ex curve,
x positive, E negative and x negative, E positive.

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