Problem

The electric field at a point due to a point charge is $26 \, \mathrm{N/C}$, and the electric potential at that point is $13 \, \mathrm{J/C}$. Calculate the distance of the point from the charge and the magnitude of the charge.

Data

• Electric Field: $E = 26 \, \mathrm{N/C}$
• Electric Potential: $V = 13 \, \mathrm{J/C}$

Prerequisite Concepts

• Electric Potential:

$$V = \frac{kq}{r}$$

• Electric Field:

$$E = \frac{kq}{r^2}$$

Answer

Step 1: Calculate Distance $r$

Using the relation $V = E \cdot r$:

$$r = \frac{V}{E}$$

Substitute values:

$$r = \frac{13}{26} = 0.5 \, \mathrm{m}$$

Step 2: Calculate Charge $q$

Using the formula $V = \frac{kq}{r}$:

$$q = \frac{V \cdot r}{k}$$

Substitute values:

$$q = \frac{(13)(0.5)}{9 \times 10^9}$$ $$q = \frac{6.5}{9 \times 10^9} = 7.22 \times 10^{-10} \, \mathrm{C}$$

Final Answer

• Distance: $r = 0.5 \, \mathrm{m}$
• Charge: $q = 7.22 \times 10^{-10} \, \mathrm{C}$

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