Force of attraction between two point charges $Q$ and $- Q$ separated by $d$ meter is $F_{e}$ . When these charges are placed two identical sphere of radius $R = 0.3 d$ whose centries are $d$ meter apart the force of attraction between them is

Greater than

F_{e}

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Equal to

F_{e}

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Less than

F_{e}

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None of these

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Solution

The correct option is B Equal to $F_{e}$
Force of attraction between two point charges $Q$ and $- Q$ separated by $d$ .
Force $F_{e}$
radius $= R = 3 d$
That is also equal to $F_{e}$ because the distance is same hence force is also same.

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Force of attraction between two point charges Q and −Q separated by d meter is Fe. When these charges are placed two identical sphere of radius R=0.3 d whose centries are d meter apart the force of attraction between them is

The correct option is B Equal to FeForce of attraction between two point charges Q and −Q separated by d.Force Feradius =R=3dThat is also equal to Fe because the distance is same hence force is also same.

Force of attraction between two point charges $Q$ and $- Q$ separated by $d$ meter is $F_{e}$ . When these charges are placed two identical sphere of radius $R = 0.3 d$ whose centries are $d$ meter apart the force of attraction between them is

The correct option is B Equal to $F_{e}$
Force of attraction between two point charges $Q$ and $- Q$ separated by $d$ .
Force $F_{e}$
radius $= R = 3 d$
That is also equal to $F_{e}$ because the distance is same hence force is also same.