ABC is a right angled triangle in which AB-3 cm and BC-4 cm - And - ABC - -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

Net force on B F_net=√(F^2_A+F_C^2) F_A=15× 12/(3)^2=20 dyne, F_C=12× 20/(4)^2=15 dyne ⇒ F_net=√(F^2_A+F_C^2)=√((20)^2+(15)^2)=25 dyne ...

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Standard XIII

Physics

Principle of Superposition

ABC is a righ...

Question

ABC is a right angled triangle in which AB=3 cm and BC=4 cm . And $∠ A B C = \frac{π}{2}$ . The three charges +15, +12 and -20 e.s.u. are placed respectively on A, B and C. The force acting on B is

125 dynes

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35 dynes

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25 dynes

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Solution

The correct option is C 25 dynes
Net force on B $F_{n e t} = \sqrt{F_{A}^{2} + F_{C}^{2}}$

$F_{A} = \frac{15 \times 12}{(3)^{2}} = 20 d y n e, F_{C} = \frac{12 \times 20}{(4)^{2}} = 15 d y n e$
$\Rightarrow F_{n e t} = \sqrt{F_{A}^{2} + F_{C}^{2}} = \sqrt{(20)^{2} + (15)^{2}} = 25 d y n e$

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ABC is a right angled triangle in which AB=3 cm and BC=4 cm . And ∠ABC=π2. The three charges +15, +12 and -20 e.s.u. are placed respectively on A, B and C. The force acting on B is

The correct option is C 25 dynesNet force on B Fnet=√F2A+F2C FA=15×12(3)2=20 dyne,FC=12×20(4)2=15 dyne ⇒Fnet=√F2A+F2C=√(20)2+(15)2=25 dyne

ABC is a right angled triangle in which AB=3 cm and BC=4 cm . And $∠ A B C = \frac{π}{2}$ . The three charges +15, +12 and -20 e.s.u. are placed respectively on A, B and C. The force acting on B is

The correct option is C 25 dynes
Net force on B $F_{n e t} = \sqrt{F_{A}^{2} + F_{C}^{2}}$

$F_{A} = \frac{15 \times 12}{(3)^{2}} = 20 d y n e, F_{C} = \frac{12 \times 20}{(4)^{2}} = 15 d y n e$
$\Rightarrow F_{n e t} = \sqrt{F_{A}^{2} + F_{C}^{2}} = \sqrt{(20)^{2} + (15)^{2}} = 25 d y n e$