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Electric Charges And Fields

Question

Two charges of + 10 μC and +20 μC are separated by a distance 2 cm. The net potential (electric) due to the pair at the middle point of the line joining the two changes, is

27 MV

18 MV

20 MV

23 MV

Solution

27 MV

Using the equation

V = $\frac{Q}{4 {πε}_{0} r}$

Q - charge

ε₀ - permittivity of free surface

The potential due to + 10 µC is

V₁ = $\frac{(10 \times 10^{- 6}) \times (9 \times 10^{9})}{1 \times 10^{- 2}}$

V₁ = 9MV

The potential due to + 20 µC is

V₂ = $\frac{20 \times 10^{- 6} \times 9 \times 10^{9}}{1 \times 10^{- 2}}$

V₂ = 18 MV

The net potential at the given point is

9 MV + 18 MV = 27 MV