1.

Please provide answer with SUITABLE 'STEPS'

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Answer:

First of all we should understand that that at the location of Q , the electric field E is GIVEN only by -2Q . Similarly at the location of 2Q , tge electric Field will be given only by Q.

Refer to the diagram for better understanding.

At the location of Q , we can say that :

| \vec E| = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{2Q}{ {d}^{2} } \bigg \}

The vector will be directed rightwards (towards 2Q).

Similarly at Location of 2Q , we can say that :

| \vec E_{2}| = \dfrac{1}{4\pi \epsilon_{0}} \bigg \{ \dfrac{Q}{ {d}^{2} } \bigg \}

= > | \vec E_{2}| = \dfrac{1}{2} \bigg\{ | \vec E| \bigg\}

Since the Q is a positive charge , the electric field at the location of -2Q will also be directed rightwards (away from Q).

Correct OPTION :

\boxed{ \red{ \bold{ \huge{ \sf{Option \: 1)}}}}}


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