Three charges each 20μ C are placed at the corners of an equilateral triangle of side 0.4m The potential energy of the.system is

Question

Three charges each 20μ C are placed at the corners of an equilateral triangle of side 0.4m The potential energy of the.system is (A) 18 × 10-6 J (B) 9 J (C) 9 × 10-6 J (D) 27 J

Tardigrade Admin · Accepted Answer

P . E , u =( kq 1 q 2/ r ) Now total P.E. = P ⋅ E AB + P ⋅ E BC + P ⋅ E AC =( k (20 μ C )(20 μ C )/(0.4)) × 3 =(9 × 109 × 20 × 20 × 10-12 × 3 × 10/4) u =27 J <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/17a3166da15508c4407d4f573391ba08-.png" />

Q. Three charges each $20 μ C$ are placed at the corners of an equilateral triangle of side 0.4m The potential energy of the.system is

$18 \times 1 0^{- 6}$ J

9 J

$9 \times 1 0^{- 6}$ J

27 J

$P . E, u = \frac{k q _{1} q _{2}}{r}$
Now total P.E. $= P \cdot E_{A B} + P \cdot E_{BC} + P \cdot E_{A C}$
$= \frac{k ( 20 μ C ) ( 20 μ C )}{( 0.4 )} \times 3$
$= \frac{9 \times 1 0 ^{9} \times 20 \times 20 \times 1 0 ^{- 12} \times 3 \times 10}{4}$
$u = 27 J$

Q. Three charges each 20μC are placed at the corners of an equilateral triangle of side 0.4m The potential energy of the.system is

18×10−6 J

9 J

9×10−6 J

27 J

P.E,u=rkq1​q2​​ Now total P.E. =P⋅EAB​+P⋅EBC​+P⋅EAC​ =(0.4)k(20μC)(20μC)​×3 =49×109×20×20×10−12×3×10​ u=27J

Q. Three charges each $20 μ C$ are placed at the corners of an equilateral triangle of side 0.4m The potential energy of the.system is

$18 \times 1 0^{- 6}$ J

$9 \times 1 0^{- 6}$ J

$P . E, u = \frac{k q _{1} q _{2}}{r}$
Now total P.E. $= P \cdot E_{A B} + P \cdot E_{BC} + P \cdot E_{A C}$
$= \frac{k ( 20 μ C ) ( 20 μ C )}{( 0.4 )} \times 3$
$= \frac{9 \times 1 0 ^{9} \times 20 \times 20 \times 1 0 ^{- 12} \times 3 \times 10}{4}$
$u = 27 J$