Question

If charges Q₁=25nC is located at (0,0) and charge Q₂=-15nC is located at (2m,0). (a) Find the resultant force on a third charge Q₀=20nC located at (2m,2m), and (b) find the electric field due to the three charges at the point (1m,1m).

Answer 1

Force on Q0 due to Q1 :

F1 = kQ1Q0/r^2

F1 = 562.5 * 10^-9N

Force on Q0 due to Q2:

F2 = kQ2Qo/r^2

F2 = -675*10^-9N

We need to find the resultant vector.

First resolve F1 into components

F1x = F1cos(45) = 399.87 nN

F1y = F1sin(45) = 399.87nN

resolve F2 into components, which has only y component:

F2x = 0

F2y = -675nN

The resultant vector for F1 an F2 is :

Fx = F1x + F2x = 399.87nN

Fy = F1y + F2y = 399.87 -675 = 275.13nN

resultant F = sqrt(Fx^2 + Fy^2)

resultant F = 485.38nN

to find angle, take the tan(angle) = Fy/Fx

so, angle = -34.9deg

F = 485.38nN, -34.9deg

b) To find the electric field at point E(1,1)

i) Find E due to Q1 :

E1 = kQ1/r^2 = 9 * 10^9 * 25 * 10^-9/2

E1 = 112.5N/C

ii) E due to Q2:

E2 = kQ2/r^2 = 9*10^9 * -15n/2 = -67.5N/C

E2 = -67.5N/C

iii) E due to Q0:

E0 = kQ0/r^2 = 9*10^9 * 20n/2 = 90N/C

The directions are as shown in the figure.

Resolve all the E vectors to its components:

E1x = E1cos(45) = 79.55

E1y = E1Sin(45) = 79.55

E2x = E2Cos(45) = 47.73

E2y = E2Sin(45) = -47.73

E0x = E0Cos(45) = -63.64

E0y = E0Sin(45) = -63.64

Find resultant E,

Ex = E1x + E2x + E0x = 63.64

Ey = E1y + E2y+E3y = -31.82

so,

E = sqrt (Ex^2 + Ey^2)

E = 37.88N/C

direction of E is :

angle = tan-1(Ey/Ex)

angle = 26.56deg

Resultant E at (1,1) is 37.88N/C ,26.56deg