Three point charges 2μC, -3μ C and -3μC are kept at the vertices A, B and C respectively of an equilateral triangle of side 20 cm. What should be the sign and magnitude of charge to be placed at the mid-point M of side BC so that charge at A remains in equilibrium?
Answers
Explanation:
Answer:
The resultant force on the charge at a is 180 N.
Explanation:
Given that,
First charge q_{1}=5\times10^{-6}\ Cq
1
=5×10
−6
C
Second charge q_{2}=10\times10^{-6}\ Cq
2
=10×10
−6
C
Third charge q_{3}=-10\times10^{-6}\ Cq
3
=−10×10
−6
C
The formula of force is defined as:
F=\dfrac{kq_{1}q_{2}}{r^2}F=
r
2
kq
1
q
2
The force on the charge at a by the charge at b
F_{ba}=\dfrac{9\times10^{9}\times5\times10^{-6}\times10\times10^{-6}}{(5\times10^{-2})^2}F
ba
=
(5×10
−2
)
2
9×10
9
×5×10
−6
×10×10
−6
F_{ba}=180\ NF
ba
=180 N
The force on the charge at a by the charge at c
F_{ac}=\dfrac{9\times10^{9}\times5\times10^{-6}\times10\times10^{-6}}{(5\times10^{-2})^2}F
ac
=
(5×10
−2
)
2
9×10
9
×5×10
−6
×10×10
−6
F_{ac}=-180\ NF
ac
=−180 N
Negative sign shows the attraction force.
The resultant force on the charge at a
F_{a}=\sqrt{(180)^2+(180)^2+2\times180\times180\times\dfrac{-1}{2}}F
a
=
(180)
2
+(180)
2
+2×180×180×
2
−1
F_{a}=180\ NF
a
=180 N
Hence, The resultant force on the charge at a is 180 N.