Three charges 1μ C, 1μ C and 2μ C are kept at vertices of A, B and C of an equilateral triangle ABC of 10 cm side respectively. The resultant force on the charge at C is

Question

Three charges 1μ C, 1μ C and 2μ C are kept at vertices of A, B and C of an equilateral triangle ABC of 10 cm side respectively. The resultant force on the charge at C is (A) 0.9 N (B) 1.8 N (C) 2.72 N (D) 3.12 N

Tardigrade Admin · Accepted Answer

The situation is shown in figure. Force on

C

due to

A

F_{A C} = \frac{1}{4 π ε _{0}} \frac{q _{A} q _{C}}{r _{A C}^{2}}

= \frac{9 \times 1 0 ^{9} \times 1 \times 2 \times 1 0 ^{- 12}}{( 0.1 ) ^{2}} = 1.8 N

Similarly,

F_{BC} = 1.8 N

Hence, net force on

C = F_{A C} + F_{BC} = 3.6 N

Q. Three charges 1μC, 1μC and 2μC are kept at vertices of A,B and C of an equilateral triangle ABC of 10cm side respectively. The resultant force on the charge at C is

0.9 N

1.8 N

2.72 N

3.12 N

The situation is shown in figure. Force on C due to A FAC​=4πε0​1​rAC2​qA​qC​​ =(0.1)29×109×1×2×10−12​=1.8N Similarly, FBC​=1.8N Hence, net force on C=FAC​+FBC​=3.6N

Q. Three charges $1 μ C,$ $1 μ C$ and $2 μ C$ are kept at vertices of $A, B$ and $C$ of an equilateral triangle $A BC$ of $10 c m$ side respectively. The resultant force on the charge at $C$ is

The situation is shown in figure. Force on $C$ due to $A$

$F_{A C} = \frac{1}{4 π ε _{0}} \frac{q _{A} q _{C}}{r _{A C}^{2}}$
$= \frac{9 \times 1 0 ^{9} \times 1 \times 2 \times 1 0 ^{- 12}}{( 0.1 ) ^{2}} = 1.8 N$
Similarly, $F_{BC} = 1.8 N$
Hence, net force on
$C = F_{A C} + F_{BC} = 3.6 N$