A point charge of $+ 2, μ C$ is placed at the origin. What is the magnitude of the electric field at a point that lies $0.15; m$ away from the charge?

easy 2 marks Class 12 · Physics · Electric Charges and Fields

Answer

8.0×10^5 N/C

The correct answer is 8.0×10^5 N/C — Divide by $r^{2}$ : $E = 1.8 \times 1 0^{4} /0.0225 = 8.0 \times 1 0^{5}; N/C$ .

Step-by-step solution

Use the formula for the electric field of a point charge: $E = \frac{1}{4 p i v a r e p s i l o n _{0}} \frac{q}{r ^{2}}$ .
Substitute $q = 2 \times 1 0^{- 6}; C$ and $r = 0.15; m$ into the formula.
Calculate $r^{2} = (0.15)^{2} = 0.0225; m^{2}$ .
Compute the numerator: $k q = 9.0 \times 1 0^{9}; N\cdotm^{2} / C^{2} \times 2 \times 1 0^{- 6}; C = 1.8 \times 1 0^{4}; N\cdotm^{2} / C$ .
Divide by $r^{2}$ : $E = 1.8 \times 1 0^{4} /0.0225 = 8.0 \times 1 0^{5}; N/C$ .

Common mistake: Using

r

instead of

r^{2}

in the denominator or forgetting the unit conversion of microcoulomb to coulomb

Concept tested: Electric field of a point charge