Question

Capacitance of a sphere A is 4 pF while of another sphere
B, it is 8 microF. If on sphere A, there is a charge 2 uc and on
B. the charge is 4 microc, then the ratio of potential of the
spheres is
6:1
1:1
4:1
2:1​

Answer 1

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Q = CV

➛Capacitance of a sphere A is 4 pF

➛ there is a charge 2 μc

➛2 x 10⁻⁶ = 4 x 10⁻¹² x V₁

V₁ = 10⁶/2

➛Capacitance of a sphere B is 8 pF

➛ there is a charge 4 μc

4 x 10⁻⁶ = 8 x 10⁻¹² x V₂

➛ V₂ = 10⁶/2

➛Hence V₁  : V₂   =  1 : 1

$\sf\pink{hope \: this \: helps \: you!! \: }$

Answer 2

Answer:

The correct answer is option (b) 1 : 1.

Concept:

Capacitance: The capacitance of a conductor is a measure of its ability to hold an electric charge.

Suppose, if a charge Q raises the potential of a conductor by V, then

Q $\propto$ V

Q = CV

where C is the capacitance of the conductor.

C = Q/V

Thus, the capacitance of a conductor is defined as the ratio of the charge given to the rise in the potential of the conductor.

Given:

Capacitance of sphere A, C₁ = 4 pF = 4 × 10⁻¹² F

Capacitance of sphere B, C₂ = 8 pF = 8 × 10⁻¹² F

Charge on sphere A, Q₁ = 2 μC = 2 × 10⁻⁶ C

Charge on sphere B, Q₂ = 4 μC = 4 × 10⁻⁶ C

Find:

The ratio of the potential of the spheres.

Solution:

From the formula, C = Q/V, we get

V = Q/C

For sphere A, let the potential be V₁, which is given by

V₁ = Q₁/C₁

    = (2 × 10⁻⁶ C) / (4 × 10⁻¹² F)

    = 0.5 × 10⁻⁶⁺¹² V

    = 0.5 × 10⁶ V

For sphere B, let the potential be V₂, which is given by

V₂ = Q₂/C₂

    = (4 × 10⁻⁶ C) / (8 × 10⁻¹² F)

    = 0.5 × 10⁻⁶⁺¹² V

    = 0.5 × 10⁶ V

Required ratio, V₁ : V₂ = V₁/V₂

                                     = (0.5 × 10⁶ V) / (0.5 × 10⁶ V)

                                     = 1/1

                                     = 1 : 1

Hence, the ratio of potential of the spheres is 1 : 1.

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