Question

What are the strength and direction of the electric field at the position indicated by the dot in the figure(Figure 1)? 

Part A

Give your answer in component form. (Assume that x-axis is horisontal and points to the right, and y-axis points upward.) 

Part B 

Give your answer as a magnitude and angle measured cw from the positive x-axis. 

Part C 

Express your answer using two significant figures ΑΣφ A2op CW from the axis Submit Request Answer

Answer 1

given,

Suppose Charge q1 = 5nC = 5*10^-9 C

q2 = 5nC = -5*10^-9 C

q3 = 10nC = 10*10^-9C

r1 = distance between q1 and dot = 4 cm = 0.04 m

r2 = distance between q2 and dot = 2 cm = 0.02 m

r3 = distance between q3 and dot

r3 = sqrt [(4cm)^2+(2cm)^2] = 4.47 cm = 0.0447 m

= 26.6 deg

Because, tan() = r1/r2

= arctan(2/4) = 26.6 deg

Now Since Electric field is given by,

E = k*q/r^2

where, k = 9*10^9

q = charge that produce electric field

r = distance from point

So, electric field at dot due to q1

E1 = k*q1/r1^2

E1= [9*10^9*5*10^-9]/(2*10^-2)^2

E1 = 11.25*10^4 N/C along +ve x-axis.

Electric field at dot due to q2

E2 = k*q2/r2^2

E2 = [9*10^9*5*10^-9]/(4*10^-2)^2

E2 = 2.8125*10^4N/C along +ve y-axis.

Electric field at dot due to q3

E3 = k*q3/r3^2

E3= [9*10^9*10*10^-9]/(4.47*10^-2)^2

E3 = 4.5*10^4 N/C at 26.6 with -ve y-axis.

Now x component of net electric field

Ex = E1+E3x

Ex = 11.25*10^4 N/C + 4.50*10^4 N/C*sin(26.60 deg)

Ex = 13.27*10^4 N/C

y component of net electric field

Ey = E2 − E3y

Ey = 2.8125*10^4 N/C − 4.50*10^4N/C*cos(26.60 deg)

Ey = -1.21*10^4N/C

part A)

So, net electric field(E) = Ex i + Ey j

E = 13*10^4 i - 1.2*10^4 j

Part-B)

Magnitude of net electric field

E = sqrt [Ex^2+Ey^2]

E = sqrt[(13.27*10^4)^2 + (-1.21*10^4)^2]

E = 13.32*10^4 N/C

Part-C)

Direction

= arctan(Ey/Ex)

= arctan((-1.21*10^4)/(13.27*10^4)) = -5.21 deg

= -5.2 deg below x-axis

Direction = 5.2 deg CW from the +x-axis

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