Question

Location C is 0.02 m from a small sphere which has a charge of 3 nanocoulombs uniformly distributed on its surface. Location D is 0.06 m from the sphere. What is the change in potential along a path from C to D?

Answer 1

The change in potential along a path from C to D due to a small charged sphere is 900 V.

Given:

Charge, Q = 3 nC = 3 × 10⁻⁹ C

Distance between the sphere and point C, r₁ = 0.02 m

Distance between the sphere and point D, r₂ = 0.06 m

Calculation:

We know that the electric potential is given as:

V = k Q/r   - (1)

where, V is the electric potential

            k is Coulomb's force constant

            Qis the charge on the  sphere

            r is the  separation distance

The electric potential at point C due to charged sphere can be given as:

V₁ = k Q/r₁

   = (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.02 m)]

   = 1350 V

The electric potential at point D due to charged sphere can be given as:

V₂ = k Q/r₂

   = (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.06 m)]

   = 450 V

Now, the change in potential along the path from C to D can be calculated as:

ΔV = V₂ - V₁

     = 450 V - 1350 V

     = -900 V

The negative sign indicates that the work is done against the electric field in moving the charge from C to D.

Therefore, the change in potential along a path from C to D is 900 V against the direction of the electric field.

Learn more about the electric potential here:

brainly.com/question/12645463

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Answer 2

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