Question

the ball in the figure are charge to 6.5 nC. whatare the strength and direction of the electric field at the position indicated by the dot in the figure?

Answer 1

Let there be angle a between field E1 and E2

E1 = K*Q1/R1
R1 = 5 cm = 0.05 m
E1 = K*Q1/R1^2
= (9*10^9)*(6.5*10^-9) / (0.05)^2
= 23400 N/C

E2 = K*Q2/R2
R2 = sqrt (5^2 + 10^2 ) cm =11.2 cm = 0.112 m
E2 = K*Q2/R2^2
= (9*10^9)*(6.5*10^-9) / (0.112)^2
= 4663.59 N/C

angle a = atan (10/5) = 63.4 degree

writing both electric field in vector form ,
E1 = 23400 x N/C
E2= 4663.59* cos a x + 4663.59 * sin a y N/C
= 4663.59* cos 63.4 x + 4663.59 * sin 63.4 y N/C
= 2088.16 x + 4169.87 y N/C

Net E = E1 + E2
= 23400 x + 2088.16x + 4169.87 y
= 25488.16 x + 4169.87 y

A)
Strength = sqrt (25488.16 ^2 + 4169.87 ^2 )
=25827.003 N/C

B)
angle from horizontal = atan (4169.87 / 25488.16)
= atan (0.164)
= 9.3 degree