Question

Electric field intensity and electric potential at a point due to a point charge are 600 V/m and -3600 V respectively. The distance of the point from the charge and the magnitude of the charge are:

• A. \(7 \, {m}, 3 \, \mu{C}\)
• B. \(8 \, {m}, 4 \, \mu{C}\)
• C. \(6 \, {m}, 2.4 \, \mu{C}\)
• D. \(4 \, {m}, 6 \, \mu{C}\)

Hint:

For electrostatic problems involving a point charge, knowing either the electric field or potential at a distance can determine the other, along with the charge's magnitude and distance.

Answer 1

The Correct Option is C

Step 1: Use the formulas for electric field and potential. \[ E = \frac{kQ}{r^2} \] \[ V = \frac{kQ}{r} \] \[ E = \frac{V}{r} \] \[ 600 = \frac{3600}{r} \] \[ r = 6 \, {m} \] Step 2: Calculate the charge \(Q\). \[ Q = \frac{r^2 E}{k} \] \[ Q = \frac{6^2 \times 600}{9 \times 10^9} \] \[ Q = 2.4 \, \mu{C} \]