Question

A charged cork of mass m suspended by a light string is placed in uniform electric filed of strength

$E=(\hat{i}+\hat{j})\times10^5 NC^{-1}$ as shown in the fig. If in equilibrium position tension in the string is $\frac{2mg}{(1+\sqrt{3})}$ then angle '$\alpha$' with the vertical is -

• A. 60°
• B. 30°
• C. 45°
• D. 18°

Answer 1

Answer: (A)

60°

Answer 2

The forces acting on the charged cork are the gravitational force $\vec{F}_g = -mg\hat{j}$, the electric force $\vec{F}_e = q\vec{E}$, and the tension in the string $\vec{T}$. In equilibrium, the vector sum of these forces is zero: $\vec{T} + \vec{F}_g + \vec{F}_e = \vec{0}$.

Let's set up a coordinate system with the vertical direction as the y-axis (upwards positive) and the horizontal direction as the x-axis (to the right positive). The electric field is given as $\vec{E} = (\hat{i} + \hat{j}) \times 10^5 \, \text{NC}^{-1}$. Let $E_0 = 10^5 \, \text{NC}^{-1}$. So $\vec{E} = E_0\hat{i} + E_0\hat{j}$. The gravitational force is $\vec{F}_g = -mg\hat{j}$. The electric force is $\vec{F}_e = q\vec{E} = qE_0\hat{i} + qE_0\hat{j}$.

The string makes an angle $\alpha$ with the vertical. Since the electric force has components in the positive x and positive y directions, the cork will be displaced to the right and upwards from the equilibrium position without the electric field. The figure shows the angle $\alpha$ with the vertical, and the string is displaced to the right. Assuming $\alpha$ is the angle with the upward vertical (positive y-axis), the tension vector can be written as $\vec{T} = T\sin\alpha\hat{i} + T\cos\alpha\hat{j}$.

The equilibrium condition is: $(T\sin\alpha\hat{i} + T\cos\alpha\hat{j}) + (-mg\hat{j}) + (qE_0\hat{i} + qE_0\hat{j}) = \vec{0}$

Separating into components: x-component: $T\sin\alpha + qE_0 = 0 \implies T\sin\alpha = -qE_0$ y-component: $T\cos\alpha - mg + qE_0 = 0 \implies T\cos\alpha = mg - qE_0$

From the figure, the string is in the first quadrant, so $\sin\alpha > 0$ and $\cos\alpha > 0$ for $0 < \alpha < 90^\circ$. Since $T > 0$ and $E_0 > 0$, the equation $T\sin\alpha = -qE_0$ implies that $-q > 0$, so $q < 0$. The charge of the cork is negative.

We are given the tension $T = \frac{2mg}{(1+\sqrt{3})}$. We can square and add the component equations: $(T\sin\alpha)^2 + (T\cos\alpha)^2 = (-qE_0)^2 + (mg - qE_0)^2$ $T^2(\sin^2\alpha + \cos^2\alpha) = q^2E_0^2 + m^2g^2 - 2mgqE_0 + q^2E_0^2$ $T^2 = 2q^2E_0^2 - 2mgqE_0 + m^2g^2$

Substitute the value of $T$: $\left(\frac{2mg}{1+\sqrt{3}}\right)^2 = 2q^2E_0^2 - 2mgqE_0 + m^2g^2$ $\frac{4m^2g^2}{(1+\sqrt{3})^2} = 2q^2E_0^2 - 2mgqE_0 + m^2g^2$ $\frac{4m^2g^2}{1 + 3 + 2\sqrt{3}} = \frac{4m^2g^2}{4 + 2\sqrt{3}} = \frac{2m^2g^2}{2 + \sqrt{3}}$ $\frac{2m^2g^2}{2 + \sqrt{3}} = 2q^2E_0^2 - 2mgqE_0 + m^2g^2$

Let's express $qE_0$ in terms of $T$ and angles. From the component equations: $-qE_0 = T\sin\alpha \implies qE_0 = -T\sin\alpha$ $mg - qE_0 = T\cos\alpha \implies mg - (-T\sin\alpha) = T\cos\alpha \implies mg + T\sin\alpha = T\cos\alpha$ $mg = T\cos\alpha - T\sin\alpha = T(\cos\alpha - \sin\alpha)$ Substitute the value of $T$: $mg = \frac{2mg}{1+\sqrt{3}}(\cos\alpha - \sin\alpha)$ $1 = \frac{2}{1+\sqrt{3}}(\cos\alpha - \sin\alpha)$ $\frac{1+\sqrt{3}}{2} = \cos\alpha - \sin\alpha$

Let's check the direction of displacement in the figure again. The electric field is in the first quadrant. Gravity is downwards. A positive charge would be pushed in the first quadrant by the electric field. A negative charge would be pushed in the third quadrant. The figure shows the cork displaced to the right and upwards. This suggests that the net force (excluding tension) is in the first quadrant, which would be the case for a positive charge. If $q > 0$, then $T\sin\alpha = -qE_0$ implies $\sin\alpha < 0$, which is not possible for $0 < \alpha < 90^\circ$. Let's re-examine the figure. The figure shows the angle $\alpha$ with the vertical line. The string is directed upwards and to the left from the cork. This means the horizontal component of tension is to the left (negative x-direction) and the vertical component is upwards (positive y-direction). If the string makes an angle $\alpha$ with the upward vertical (positive y-axis) and is in the second quadrant, then $T_x = -T\sin\alpha$ and $T_y = T\cos\alpha$. Equilibrium equations: x-component: $-T\sin\alpha + qE_0 = 0 \implies T\sin\alpha = qE_0$ y-component: $T\cos\alpha - mg + qE_0 = 0 \implies T\cos\alpha = mg - qE_0$

From $T\sin\alpha = qE_0$, since $T > 0$, $\sin\alpha > 0$ (for $0 < \alpha < 90^\circ$), and $E_0 > 0$, we must have $q > 0$. From $T\cos\alpha = mg - qE_0$, since $T > 0$ and $\cos\alpha > 0$, we need $mg - qE_0 > 0 \implies mg > qE_0$. So, $mg > T\sin\alpha$.

Divide the component equations: $\frac{T\sin\alpha}{T\cos\alpha} = \frac{qE_0}{mg - qE_0}$ $\tan\alpha = \frac{qE_0}{mg - qE_0}$

From the equations $T\sin\alpha = qE_0$ and $T\cos\alpha = mg - qE_0$, we have $qE_0 = T\sin\alpha$. $T\cos\alpha = mg - T\sin\alpha$ $mg = T\cos\alpha + T\sin\alpha = T(\cos\alpha + \sin\alpha)$ Substitute the value of $T$: $mg = \frac{2mg}{1+\sqrt{3}}(\cos\alpha + \sin\alpha)$ $1 = \frac{2}{1+\sqrt{3}}(\cos\alpha + \sin\alpha)$ $\frac{1+\sqrt{3}}{2} = \cos\alpha + \sin\alpha$

$\alpha = 60^\circ$: $\cos 60^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}$.

The angle $\alpha$ with the vertical is $60^\circ$.