Electrostatics Question 178
Question: $ ABC $ is a right angled triangle in which $ AB=3cm $ and $ BC=4cm $ . And Ð ABC = p/2. The three charges $ +15,\ +12 $ and $ -20e.s.u. $ are placed respectively on $ A $ , $ B $ and $ C $ . The force acting on $ B $ is
Options:
A) $ 125\ dynes $
B) $ 35\ dynes $
C) $ 25\ dynes $
D) Zero
Show Answer
Answer:
Correct Answer: C
Solution:
Net force on B $ F _{net}=\sqrt{F _{A}^{2}+F _{C}^{2}} $
$ F _{A}=\frac{15\times 12}{{{( 3 )}^{2}}}=20dyne $ , $ F _{C}=\frac{12\times 20}{{{( 4 )}^{2}}}=15dyne $
therefore $ F _{net}=\sqrt{F _{A}^{2}+F _{C}^{2}}=\sqrt{{{(20)}^{2}}+{{(15)}^{2}}}=25dyne $
BETA
