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A negative charge is placed at the midpoint between two fixed equal positive charges , separated by a distance 2d . If the negative charge is given a small distancement x(xltltd) perpendicular to the line joining the positive charges , how the force (F) developed on it will approximately depend on x ? |
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Answer» `Fpropx`  Let us consider that , TWO positive CHARGES be `+Q` and negative charge be -q `AC=BC=r=(x^(2)+d^(2))^(1//2)` … (i) Now , electric field at point C due to the charges +Q placed at A and B is `E_("net")=2Ecostheta=2xx(kQ)/(r^(2))xx(x)/(r)=(2kQx)/(r^(3))` `therefore` Force on charge - q will be , `F=-qE_(net)` `=-(2kQqx)/(r^(3))=-(2kQqx)/((x^(2)+d^(2))^(3//2))`(using (i)) Now , for small displacement `(x),xltltd,x^(2)=0` `thereforeF=-(2kQqx)/(d^(3))rArrFpropx` So , the negative charge will OSCILLATE along its mean POSITION and the force acting on the negative charge will be restoring force. |
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