(a) What is the electric potential due to the 1.0 µC charge at the third vertex,
ID: 1573474 • Letter: #
Question
(a) What is the electric potential due to the 1.0 µC charge at the third vertex, point P?
(b) What is the electric potential due to the -2.3 µC charge at P?
(c) Find the total electric potential at P.
(d) What is the work required to move a 3.8 µC charge from infinity to point P?
Explanation / Answer
Given
charges are q1 = 1.0 µC and q2 = -2.3 µC are d = 0.42 m apart at two vertices of an equilateral triangle
we know that the potential at a point due to a point q at a distance r is
V = k*q/r
a)the electric potential due to the 1.0 µC charge at the third vertex, point P is
v1 = k*q1/d
v1 = 9*10^9*1*10^-6/0.42 V
V1 = 21428.57 V
b)the electric potential due to the -2.3 µC charge at the third vertex, point P is
v1 = k*q2/d
v1 = 9*10^9*(-2.3)*10^-6/0.42 V
V1 = - 49285.71 V
c) the total electric potential at P is
V_P = V1+V2 = 21428.57- 49285.71 V = -27857.14 V
d) the work required to move a 3.8 µC charge from infinity to point P is
by definition of the potential is
the workdone on a unit postive test charge to bring it from infinity to a point inside the electric field.
that is V = W/q
now to move 3.8 µC charge from infinity to point P , the work required is
W = V*q = -27857.14 *3.8*10^-6 J = -0.105857132 J
so work required is W = 0.105857132 J
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