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

Question

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

Express your answer in terms of the unit vectors i^i and jj^. Express your answer using two significant figures.

E=

B)Give your answer as a magnitude and angle measured cw from the positive x

x-axis.

Express your answer using two significant figures.

E=

C)Express your answer using two significant figures.

0 =

Answer 1

Step 1:

(a)

Observe the figure:

The three point charges are , and

The angle of the field at the dot from the bottom charge() is .

The angle of the field at the dot from the top charge() is .

The angle of the field at the dot from the top left charge() is .

Electric field due to charged particle is , where

 is the permittivity of free space .

 is the charge of the electron.

 is the distance between the two point charges.

Step 2:

The distance between the bottom charge and the dot is .

For the charged particle along bottom surface is .

The distance between the top charge and the dot is .

For the charged particle along bottom surface is .

The distance between the top left charge and the dot is .

For the charged particle along bottom surface is .

Step 3:

Find the net electric field.

is sum of horizontal and vertical components.

Find the sum of horizontal components.

Find the sum of vertical components.

Therefore .

Solution:

.

Comments

Ex is actually 1.2x10^5 + 2.0x10^4

Ey is actually 2.8x10^4 -1.2x10^4

With correct answer E= 1.3x10^5i - 1.2x10^4j

a minor mix up of the final vectors from the top left (diag) charge.  Verified my answer on mastering physics.

Answer 2

(b)

From (a): .

Find the magnitude of electric field strength.

Strength of the electric field is

Answer 3

(c)

From (a): .

Find the direction of electric field.

Direction of electric field is 3° clockwise.