Question

Two metallic spheres of radii 1:2 are connected by a conducting wire. What is the ratio of electric field intensities at their surface?

Hint:

When two metallic spheres are connected by a conducting wire, they have the same potential, and the ratio of their electric field intensities is related to the ratio of their radii.

Answer 1

Step 1: Use the formula for electric field.
The electric field intensity \( E \) at the surface of a sphere is given by the formula: \[ E = \frac{kQ}{r^2} \] where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the radius of the sphere. Since the spheres are connected by a conducting wire, they must have the same potential, which implies that the ratio of the charges on the spheres will be proportional to the ratio of their radii.
Step 2: Relationship between charge and radius.
Let the charges on the two spheres be \( Q_1 \) and \( Q_2 \), and their radii be \( r_1 = r \) and \( r_2 = 2r \), respectively. Since the potential is the same, we have: \[ \frac{Q_1}{r_1} = \frac{Q_2}{r_2} \] This implies: \[ Q_1 = \frac{r_1}{r_2} Q_2 = \frac{1}{2} Q_2 \]
Step 3: Find the ratio of electric field intensities.
The electric field intensities at the surfaces of the spheres are: \[ E_1 = \frac{kQ_1}{r_1^2} \quad \text{and} \quad E_2 = \frac{kQ_2}{r_2^2} \] Substitute \( Q_1 = \frac{1}{2} Q_2 \) and \( r_2 = 2r \): \[ E_1 = \frac{k \cdot \frac{1}{2} Q_2}{r^2} \quad \text{and} \quad E_2 = \frac{kQ_2}{(2r)^2} = \frac{kQ_2}{4r^2} \] Now, calculate the ratio of electric field intensities: \[ \frac{E_1}{E_2} = \frac{\frac{k \cdot \frac{1}{2} Q_2}{r^2}}{\frac{kQ_2}{4r^2}} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2 \]
Step 4: Conclusion.
The ratio of the electric field intensities at the surface of the two spheres is: \[ \boxed{2} \]