Question

Electric charges of 1μC, -1 μC and 2μC are placed in air at the corners A,B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is​

Answer 1

Answer:

See the diagram.

The force on C due to A,

F

CA

=

0.1

2

9×10

9

×2×1

=1.8N

The force on C due to B F

CB

=1.8N

Doing vectorial addition of the forces using,

c

2

=a

2

+b

2

+2abcosθ=1.8

2

+1.8

2

+2×1.8

2

×cos(120)=1.8

2

Answer 2

Given: Electric charges of 1μC, -1 μC, and 2μC are placed in the air at the corners A, B, and C respectively of an equilateral triangle ABC having a length of each side 10 cm.

To find: The resultant force on the charge at C.

Solution:

The electric charge placed at point A = 1μC

The electric charge placed at point B = -1μC

The electric charge placed at point C = 2μC

Length of each side of equilateral triangle = 10 cm

Force = k × $\frac{q_{1}q_{2} }{d^{2} }$

Here, k is the Coulomb's constant whose value is equal to 9 × $10^{9}$ N$m^{2} C^{-2}$

There will be two forces on C, one due to A and another due to B.

$F_{A}$ = force on C due to charge placed at point A

So, $F_{A}$ = 9 × $10^{9}$ $\frac{(10^{-6})(2. 10^{-6}) }{(0.1)^{2} }$ = 1.8 N

Now, $F_{B}$ = force on C due to charge placed at point B

$F_{B}$ = 9 × $10^{9}$ $\frac{(10^{-6})(2. 10^{-6}) }{(0.1)^{2} }$ = 1.8 N

So, Net force on C = $\sqrt{(F_{A}) ^{2} } + (F_{B} )^{2} + 2F_{A} F_{B}cos 120$ = 1.8 N

Thus, the resultant force on the charge at C is 1.8 N.