Skip to content
Numerical · Q20

Q.Find the magnitude of the electric field at a distance of 2 m2\ \text{m} from an isolated point charge of +5 μC+5\ \mu\text{C} in air.

West Bengal WbchseTextbookSubjectiveImportance★★★★★est
48% · 20/42 Questions
✓ Free question

Given q=+5 μC=5×10−6 Cq=+5\ \mu\text{C}=5\times10^{-6}\ \text{C}, r=2 mr=2\ \text{m}.

E=kqr2=(9×109)(5×10−6)(2)2=4.5×1044=1.125×104 N/CE = \frac{kq}{r^2} = \frac{(9\times10^9)(5\times10^{-6})}{(2)^2} = \frac{4.5\times10^4}{4} = 1.125\times10^4\ \text{N/C}

✓Final answer

E=1.125×104 N/CE=1.125\times10^4\ \text{N/C} (=11250 N/C=11250\ \text{N/C}), pointing radially outward from the charge (since it is positive).

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.