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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Electrostatics

JIPMER 2015 · Physics · Electrostatics

Three equal charges, each having a magnitude of 2.0 × 10^ -6 C ,...

Three equal charges, each having a magnitude of 2.0×10−6C2.0 \times 10^{-6} \mathrm{C}, are placed at the three corners of a right angled triangle of sides 3 cm,4 cm3 \mathrm{~cm}, 4 \mathrm{~cm} and 5 cm . The force (in magnitude) on the charge at the right angled corner isThree equal charges, each having a magnitude of 2.0 × 10^ -6 C , are placed at the three corners of a right angled

Options

  1. A50 N
  2. B26 N
  3. C29 N
  4. D45.9 N

Correct answer

D. 45.9 N

Step-by-step solution

Consider the diagram Three equal charges, each having a magnitude of 2.0 × 10^ -6 C , are placed at the three corners of a right angled The situation is shown in figure. The force on AA due to BB is F1=(2.0×10−6)(2.0×10−6)4πε0(4 cm)2F_1=\frac{\left(2.0 \times 10^{-6}\right)\left(2.0 \times 10^{-6}\right)}{4 \pi \varepsilon_0(4 \mathrm{~cm})^2}=9×109×4×10−12×116×10−4=22.5 N\begin{aligned}=9 \times 10^9 \times 4 \times 10^{-12} & \times \frac{1}{16 \times 10^{-4}} \\ & =22.5 \mathrm{~N}\end{aligned}As the force due to $C$ on $A$ is 40 N . Thus, net force (electric)\begin{aligned} F & =\sqrt{F_1^2+F_1^2} \\ & =\sqrt{(22.5)^2+(40)^2}=45.9 \mathrm{~N}\end{aligned}$$
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