JEE Main & Advanced
Physics
Electrostatics & Capacitance
Question Bank
Charge and Coulombs Law
question_answer
\[ABC\] is a right angled triangle in which \[AB=3\,cm\] and \[BC=4\,cm\]. And Ð ABC = p/2. The three charges \[+15,\ +12\] and \[-20\,e.s.u.\] are placed respectively on \[A\], \[B\] and \[C\]. The force acting on \[B\] is
A) \[125\ dynes\]
B) \[35\ dynes\]
C) \[25\ dynes\]
D) Zero
Correct Answer:
C
Solution :
Net force on B \[{{F}_{net}}=\sqrt{F_{A}^{2}+F_{C}^{2}}\] \[{{F}_{A}}=\frac{15\times 12}{{{\left( 3 \right)}^{2}}}=20\,\,dyne\], \[{{F}_{C}}=\frac{12\times 20}{{{\left( 4 \right)}^{2}}}=15\,\,dyne\] Þ \[{{F}_{net}}=\sqrt{F_{A}^{2}+F_{C}^{2}}=\sqrt{{{(20)}^{2}}+{{(15)}^{2}}}=25\,\,dyne\]