A charged cork of mass m suspended by a light string is placed in uniform electric field of strength E = ( i^ + j^ ) * 10⁵ N/C. If in equilibrium position tension in string is 2mg/ ( 1 + root3), the angle ‘alpha’ with the vertical is :

To find the angle 'alpha' that the charged cork makes with the vertical when it's in equilibrium within the electric field, we can analyze the forces acting on the cork. Given that the tension in the string is \( \frac{2mg}{1 + \sqrt{3}} \), we can break it down step-by-step.

Understanding the Forces

In equilibrium, the forces acting on the cork consist of:

• The gravitational force (\( mg \)) acting downward.
• The tension (\( T \)) in the string acting along the string at an angle \( \alpha \) with the vertical.
• The electric force (\( F_e \)), which is the force due to the electric field acting horizontally.

Components of Forces

We can resolve the tension into two components:

• The vertical component: \( T \cos(\alpha) \)
• The horizontal component: \( T \sin(\alpha) \)

Equations of Motion

In the vertical direction, since the system is in equilibrium:

\( T \cos(\alpha) = mg \tag{1} \)

In the horizontal direction, the electric force acting on the cork is equal to the horizontal component of the tension:

\( F_e = T \sin(\alpha) \tag{2} \)

Calculating the Electric Force

The electric force can be calculated using the formula:

\( F_e = qE \tag{3} \)

Where \( q \) is the charge on the cork and \( E \) is the electric field strength. In this case, \( E \) is given as \( (1\hat{i} + 1\hat{j}) \times 10^5 \) N/C, which has a magnitude of:

\( E = \sqrt{(10^5)^2 + (10^5)^2} = 10^5 \sqrt{2} \, \text{N/C} \)

Substituting Values

Now, substituting the value of tension into equations (1) and (2):

From equation (1):

\( T \cos(\alpha) = mg \)

\( \frac{2mg}{1 + \sqrt{3}} \cos(\alpha) = mg \)

Dividing both sides by \( mg \):

\( \frac{2}{1 + \sqrt{3}} \cos(\alpha) = 1 \)

Thus:

\( \cos(\alpha) = \frac{1 + \sqrt{3}}{2} \tag{4} \)

Finding the Angle Alpha

Now we can find \( \alpha \) using the cosine value derived:

\( \alpha = \cos^{-1}\left(\frac{1 + \sqrt{3}}{2}\right) \)

Using a calculator or trigonometric tables, we can find the angle. The value \( \cos(\alpha) \) suggests that \( \alpha \) will be in the range of 0 to 90 degrees, which is typical for an angle made with the vertical. Evaluating this will give us the final angle of the cork with the vertical.

Final Consideration

In conclusion, you can use these steps to determine the angle that the cork makes with the vertical when it is in equilibrium in a uniform electric field. Make sure to apply the values carefully for accurate results!