Question

Two point charges Q and 16 Q are fixed at separation d . Where should a third point charge be placed between them so that it experiences no force I.e. in equilibrium ? (b) In the previous problem , if we replace Q by − Q , where should the third charge be placed on the joining the charges so that it is equilibrium ? (c ) Three point charges Q , q and Q are placed at x = 0 , x = d / 2 and x = d . Find q in terms of Q so that all charges are in equilibrium. (d) Two point charges 9 Q and 25 Q are placed at separation d . Where should a third charge q be placed between them so that all charges are in equilibrium ? Also find the value of q and its nature.

Answer 1

a) The third charge should be placed at a distance d/5 from the charge Q.

b) The third charge should be placed at a distance d/3 to the left of the charge -Q.

c) q = -Q/4.

d) Position: 3d/8 from 9Q. Nature: Negative. Value: q = -225Q/64.

Answer 2

The problem deals with electrostatic equilibrium, where the net force on a charge (or all charges) is zero. We use Coulomb's Law, $F = k \frac{|q_1 q_2|}{r^2}$, and the principle of superposition of forces.

a) Two point charges Q and 16Q are fixed at separation d. Where should a third point charge be placed between them so that it experiences no force?

Let the charges be $Q_1 = Q$ at $x=0$ and $Q_2 = 16Q$ at $x=d$. Let the third charge be $q$ placed at a distance $x$ from $Q_1$. Since both $Q_1$ and $Q_2$ are positive, for $q$ to experience no force, it must be placed between them. The forces exerted by $Q_1$ and $Q_2$ on $q$ will be in opposite directions. The force on $q$ due to $Q_1$ is $F_1 = k \frac{Q q}{x^2}$. The force on $q$ due to $Q_2$ is $F_2 = k \frac{16Q q}{(d-x)^2}$. For equilibrium, $F_1 = F_2$: $k \frac{Q q}{x^2} = k \frac{16Q q}{(d-x)^2}$ $\frac{1}{x^2} = \frac{16}{(d-x)^2}$ Taking the square root of both sides (since $x$ and $d-x$ are positive distances): $\frac{1}{x} = \frac{4}{d-x}$ $d-x = 4x$ $d = 5x$ $x = \frac{d}{5}$ The third charge should be placed at a distance $\frac{d}{5}$ from the charge $Q$ (and $\frac{4d}{5}$ from $16Q$).

b) In the previous problem, if we replace Q by -Q, where should the third charge be placed on the joining the charges so that it is equilibrium?

Now, the charges are $Q_1 = -Q$ at $x=0$ and $Q_2 = 16Q$ at $x=d$. For a third charge $q$ to be in equilibrium, the forces on it due to $Q_1$ and $Q_2$ must be equal in magnitude and opposite in direction. If $q$ is placed between $-Q$ and $16Q$, the forces from them would be in the same direction (e.g., if $q$ is positive, $-Q$ attracts it to the left, and $16Q$ repels it to the left; if $q$ is negative, $-Q$ repels it to the right, and $16Q$ attracts it to the right). Thus, equilibrium is not possible between them. The equilibrium position must be outside the region between the charges, and closer to the charge with smaller magnitude. Here, $|-Q| < |16Q|$, so the equilibrium point is to the left of $-Q$. Let the third charge $q$ be placed at a distance $y$ to the left of $-Q$. So its position is $x=-y$. The distance from $-Q$ is $y$. The distance from $16Q$ is $d+y$. The force on $q$ due to $-Q$ is $F_1 = k \frac{|-Q| q}{y^2} = k \frac{Q q}{y^2}$. The force on $q$ due to $16Q$ is $F_2 = k \frac{16Q q}{(d+y)^2}$. For equilibrium, $F_1 = F_2$: $k \frac{Q q}{y^2} = k \frac{16Q q}{(d+y)^2}$ $\frac{1}{y^2} = \frac{16}{(d+y)^2}$ Taking the square root of both sides: $\frac{1}{y} = \frac{4}{d+y}$ $d+y = 4y$ $d = 3y$ $y = \frac{d}{3}$ The third charge should be placed at a distance $\frac{d}{3}$ to the left of the charge $-Q$.

c) Three point charges Q, q and Q are placed at x = 0, x = d/2 and x = d. Find q in terms of Q so that all charges are in equilibrium.

Let $Q_1 = Q$ at $x=0$, $Q_2 = q$ at $x=d/2$, and $Q_3 = Q$ at $x=d$. For all charges to be in equilibrium, the net force on each charge must be zero.

1. Equilibrium of $Q_2$ (charge $q$ at $x=d/2$): The force on $q$ due to $Q_1$ is $F_{12} = k \frac{Q q}{(d/2)^2} = k \frac{4Qq}{d^2}$. The force on $q$ due to $Q_3$ is $F_{32} = k \frac{Q q}{(d/2)^2} = k \frac{4Qq}{d^2}$. For $q$ to be in equilibrium, $F_{12}$ and $F_{32}$ must be equal in magnitude and opposite in direction. Since their magnitudes are already equal, they must be opposite. This implies that $q$ must have the opposite sign to $Q$. If $Q$ is positive, $q$ must be negative (forces are attractive). If $Q$ is negative, $q$ must be positive (forces are attractive).
2. Equilibrium of $Q_1$ (charge $Q$ at $x=0$): The force on $Q_1$ due to $Q_3$ is $F_{31} = k \frac{Q Q}{d^2} = k \frac{Q^2}{d^2}$. (This force is repulsive, acting to the left). The force on $Q_1$ due to $Q_2$ ($q$) is $F_{21} = k \frac{q Q}{(d/2)^2} = k \frac{4qQ}{d^2}$. For $Q_1$ to be in equilibrium, $F_{21}$ must be equal in magnitude and opposite in direction to $F_{31}$. So $F_{21}$ must act to the right, meaning it must be attractive. This confirms that $q$ must be opposite in sign to $Q$. Equating magnitudes: $k \frac{Q^2}{d^2} = k \frac{|4qQ|}{d^2}$ $Q^2 = 4|q|Q$ $Q = 4|q|$ $|q| = \frac{Q}{4}$ Since $q$ must be opposite in sign to $Q$, we have $q = -\frac{Q}{4}$. (By symmetry, if $Q_1$ is in equilibrium, $Q_3$ will also be in equilibrium).

d) Two point charges 9Q and 25Q are placed at separation d. Where should a third charge q be placed between them so that all charges are in equilibrium? Also find the value of q and its nature.

Let $Q_1 = 9Q$ at $x=0$, $Q_2 = q$ at $x=x$, and $Q_3 = 25Q$ at $x=d$.

1. Equilibrium of $Q_2$ (charge $q$): For $q$ to be in equilibrium, it must be placed between $9Q$ and $25Q$. The force on $q$ due to $9Q$ is $F_1 = k \frac{9Q q}{x^2}$. The force on $q$ due to $25Q$ is $F_2 = k \frac{25Q q}{(d-x)^2}$. For equilibrium, $F_1 = F_2$: $k \frac{9Q q}{x^2} = k \frac{25Q q}{(d-x)^2}$ $\frac{9}{x^2} = \frac{25}{(d-x)^2}$ Taking the square root: $\frac{3}{x} = \frac{5}{d-x}$ $3(d-x) = 5x$ $3d - 3x = 5x$ $3d = 8x$ $x = \frac{3d}{8}$ So, the third charge $q$ is placed at a distance $\frac{3d}{8}$ from $9Q$. For $q$ to be in equilibrium between two positive charges ($9Q$ and $25Q$), $q$ must be negative. This is because the forces from $9Q$ and $25Q$ on $q$ must be attractive (opposite directions). So, the nature of $q$ is negative.

2. Equilibrium of $Q_1$ (charge $9Q$ at $x=0$): The force on $9Q$ due to $25Q$ is $F_{31} = k \frac{25Q \cdot 9Q}{d^2} = k \frac{225Q^2}{d^2}$. (Repulsive, acting to the left). The force on $9Q$ due to $q$ is $F_{21} = k \frac{|q| (9Q)}{x^2}$. Since $q$ is negative, this force is attractive, acting to the right. Substitute $x = \frac{3d}{8}$: $F_{21} = k \frac{|q| (9Q)}{(3d/8)^2} = k \frac{9|q|Q}{9d^2/64} = k \frac{64|q|Q}{d^2}$. For equilibrium, $F_{31} = F_{21}$: $k \frac{225Q^2}{d^2} = k \frac{64|q|Q}{d^2}$ $225Q^2 = 64|q|Q$ $225Q = 64|q|$ $|q| = \frac{225}{64}Q$ Since $q$ is negative, $q = -\frac{225}{64}Q$. (By symmetry, if $Q_1$ is in equilibrium, $Q_3$ will also be in equilibrium).