Two pint charges +3q and -4q are placed at the vertices B and C of an equilateral triangle ABC of side a units. Obtain the expression for the magnitude and direction of electric field at the vertex A due to these two charges.
Answers
Explanation:
Two point charges + q and - 2q are placed at the vertices 'B' and 'C' of an equilateral triangle ABC of side 'a' as given in the figure. Obtain the expression
Explanation:
(a) :
Forces acting on the charge q are shown in the figure.
Magnitude of force F
1
is given by F
1
=
l
2
K2q
2
Magnitude of force F
2
is given by F
2
=
l
2
K4q
2
Resolving these forces in x and y components.
x direction :
Net force F
x
=−(F
1
cos60+F
2
cos60)
x
^
⟹ F
x
=−(
l
2
k2q
2
.
2
1
+
l
2
k4q
2
.
2
1
)
x
^
=−
l
2
3kq
2
x
^
y direction :
Net force F
y
=−(F
2
sin60−F
2
sin60)
y
^
⟹ F
y
=−(
l
2
k4q
2
.
2
3
−
l
2
k2q
2
.
2
3
)
y
^
=−
l
2
3
kq
2
y
^
Thus net force on q, F
net
=F
x
+F
y
=−
l
2
kq
2
(3
x
^
+
3
y
^
)
(b) :
Potential energy of the system U=U
q,2q
+U
q,−4q
+U
2q,−4q
U=
l
kq(2q)
+
l
kq(−4q)
+
l
k2q(−4q)
=
l
−10kq
2
Thus work done to separate them to infinity W=U=
l
−10kq
2
