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Two protons $A$ and $B$ are placed in space between plates of a parallel plate capacitor charged upto $V$ volts (See fig.) Forces on protons are ${F_A}$ and ${F_B}$, then

${F_A} > {F_B}$
${F_A} < {F_B}$
${F_A} = {F_B}$
Nothing can be said
Solution
(c) ${F_A} = {F_B}$; because an uniform electric field produced between the plates.
Similar Questions
Four charge $Q _1, Q _2, Q _3$, and $Q _4$, of same magnitude are fixed along the $x$ axis at $x =-2 a – a ,+ a$ and $+2 a$, respectively. A positive charge $q$ is placed on the positive $y$ axis at a distance $b > 0$. Four options of the signs of these charges are given in List-$I$ . The direction of the forces on the charge q is given in List-$II$ Match List-$1$ with List-$II$ and select the correct answer using the code given below the lists.$Image$
List-$I$ | List-$II$ |
$P.$ $\quad Q _1, Q _2, Q _3, Q _4$, all positive | $1.\quad$ $+ x$ |
$Q.$ $\quad Q_1, Q_2$ positive $Q_3, Q_4$ negative | $2.\quad$ $-x$ |
$R.$ $\quad Q_1, Q_4$ positive $Q_2, Q_3$ negative | $3.\quad$ $+ y$ |
$S.$ $\quad Q_1, Q_3$ positive $Q_2, Q_4$ negative | $4.\quad$ $-y$ |