Point charge q1 = -5.20 nC is at the origin and point charge q2 = +3.15 nC is on

Question

Point charge q1 = -5.20 nC is at the origin and point charge q2 = +3.15 nC is on the x-axis at x = 2.85 cm. Point P is on the y-axis at y = 3.90 cm.
(a) Calculate the electric fields [E with arrow] 1 and [E with arrow] 2 at point P due to the charges q1 and q2. Express your results in terms of unit vectors.

[E with arrow] 1 =  N/C [i] +  N/C [j]
[E with arrow] 2 =  N/C [i] +  N/C [j]

(b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form.

[E with arrow] =  N/C [i] +  N/C [j]

Explanation / Answer

Here,

for the charge q1 ,

E1 = k*q1/d^2 (-j)

E1= 5.2 *10^-9 * 9*10^9/(0.039)^2 (-j)

E1 = - 3.08 *10^4 j N/C

Now, for the charge q2

theta = arctan(3.90/2.85)

theta = 53.8 degree

Now ,

E2 = (9*10^9 * 3.15 *10^-9/(0.0285^2 + 0.039^2))*(-cos(53.8) i + j * sin(53.8))

E2 = -7176.1 i + 9804.9 j N/C

b)

Now, for the net electic field at E ,

E = - 3.08 *10^4 j + -7176.1 i + 9804.9 j

E = ( -7176.1 i - 20995 j ) N/C

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