(b) : Here $P Q=1 m , E_2=1.02 V$ When $S$ is open, $P J=P Q-J Q=100-49=51 cm$ Also, $E_2=k P J$ Potential gradient in the wire, $k=\frac{E_2}{P J}$ $ k=\frac{1.02 V }{51 cm }=0.02 V / cm $
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