$A$ charge $Q$ is placed at each of the opposite corners of a square. $A$ charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero,then $\frac{Q}{q} = $ . . . . . .

  • A
    $-2 \sqrt{2}$
  • B
    $-1$
  • C
    $1$
  • D
    $-\frac{1}{\sqrt{2}}$

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Six point charges are placed at the vertices of a regular hexagon as shown in the figure. Three of the charges are $+q$ and the other three are $-q$. Starting from $P$ and moving clockwise,the electric field at the center $O$ is twice the electric field due to a single charge $+q$ at $R$. Which of the following arrangements of charges at $P, Q, R, S, T, U$ is correct?

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Four charges $Q_1, Q_2, Q_3$, and $Q_4$ of the same magnitude are fixed along the $x$-axis at $x = -2a, -a, +a$, and $+2a$, respectively. A positive charge $q$ is placed on the positive $y$-axis at a distance $b > 0$. Four options regarding the signs of these charges are given in List-$I$. The direction of the net force on the charge $q$ is given in List-$II$. Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists.
List-$I$List-$II$
$P. Q_1, Q_2, Q_3, Q_4$ all positive$1. +x$
$Q. Q_1, Q_2$ positive; $Q_3, Q_4$ negative$2. -x$
$R. Q_1, Q_4$ positive; $Q_2, Q_3$ negative$3. +y$
$S. Q_1, Q_3$ positive; $Q_2, Q_4$ negative$4. -y$

Two spheres of electric charges $+2 \ nC$ and $-8 \ nC$ are placed at a distance '$d$' apart. If they are allowed to touch each other,what is the new distance between them to get a repulsive force of the same magnitude as before?

$A$ total charge $q$ is divided into $q_1$ and $q_2$,which are placed at two vertices of an equilateral triangle of side $a$. The magnitude of the electric field $E$ at the third vertex of the triangle is to be depicted schematically as a function of $x = q_1 / q$. Choose the correct figure.

The repulsive force between two point charges is $F$ when they are separated by a distance of $1 \, m$. Now,these point charges are replaced by spheres of radius $25 \, cm$ having the same charges. The distance between their centers is $1 \, m$. The repulsive force in the two cases will decrease according to:

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