Question

A charge of ? = −5.0 × 10^ -6 ? is placed at the origin. The point P is the point (4.0 × 10^ −2 , 2.0 × 10 ^−2 ) ?. Determine:

a) the magnitude of the electric field at the point P

b) the x and y components of the electric field at the point P

c) the electric potential at the point P

Answer 1

a.)

Magnitude of the electric field is given by,

E = k*q/r^2

here, k = 9*10^9

q = magnitude of charge = 5.0*10^-6 C

r = distance of point 'P' (4.0*10^−2 , 2.0*10^−2 )m from the charge = sqrt(0.04^2 + 0.02^2) = 0.0447 m

then, E = (9*1069)*(5.0*10^-6)/0.0447^2

E = 22521508 N/C

E = 2.25*10^7 N/C

b.)

As, direction of electric field is away from positive charge and towards the negative charge.

So, direction of electric field() = arctan(0.02/0.04) = 26.56 deg above the +x direction.

then, x-component of electric field at point 'P' will be:

Ex = E*cos = (2.25*10^7)*cos(26.56 deg)

Ex = 2.01*10^7 N/C

Also, y-component of electric field at point 'P' will be:

Ey = E*sin = (2.25*10^7)*sin(26.56 deg)

Ey = 1.01*10^7 N/C

c.)

Electric potential is a scalar quantity, which is given by,

V = k*q/r

using known values:

V = (9*10^9)*(-5.0*10^-6)/0.0447

V = -1.01*10^6 N.m/C

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