Target Exam

CUET

Subject

Physics

Chapter

Electric Charges and Fields

Question:

Two point charges $q_1 = 4 μC$ & $q_2 = 9 μC$ are placed 20 cm apart. The electric field due to them will be zero on the line joining them at a distance of

Options:

8 cm from $q_1$

8 cm from $q_2$

80/13 cm from $q_1$

80/13 cm from $q_2$

Correct Answer:

8 cm from $q_1$

Explanation:

Let electric field is zero at distance x from q1

⇒ $\frac{kq_1}{x^2} = \frac{kq_2}{(20-x)^2}$

⇒ $\frac{4}{x^2} = \frac{9}{(20-x)^2}$

⇒ $\frac{2}{x} = \pm \frac{3}{20-x}$

⇒ 40-2x = $\pm$ 3x 

x = 8 cm from q