# Find the electric field at x=50cm and x=100cm?

## Two charges are placed on the x-axis: $+ 5 \mu C$ charge is placed at x=80cm and $- 4 \mu C$ charge is placed at x=0. Find E at a) x=50cm and b) x=100cm. The formula for finding the electric field is $E = \frac{k q}{r} ^ 2$ I know there will not be any y-components because the points are on the x-axis. How do I get the charge at x=50cm? In some of the previous questions I had to equate two of the electric fields, but those questions seem to be much different than this one. I think I might be overlooking something important, but I am completely lost for now. Thanks very much in advance for the help.

Jul 2, 2018

a. So the sum is $\frac{k \cdot 71.5 \cdot {10}^{-} 6 Q}{m} ^ 2$ pointing left
b. So the sum is $\frac{k \cdot 121 \cdot {10}^{-} 6 Q}{m} ^ 2$ pointing right.

#### Explanation:

This calls for recognizing that electric fields are vectors and they can be added like other vectors. So individually calculate the electric field at the specific point from both of the charges. Then add.

a. At x = 0.5 m, the field from the charge at x = 0.8 m:
${\vec{E}}_{1} = \frac{k q}{r} ^ 2 = \frac{k \cdot 5 \cdot {10}^{-} 6 Q}{0.3 m} ^ 2$
${\vec{E}}_{1} = \frac{k \cdot 11.1 \cdot 5 \cdot {10}^{-} 6 Q}{m} ^ 2 = \frac{k \cdot 55.5 \cdot {10}^{-} 6 Q}{m} ^ 2$
This will be to the left at x = 0.5 m.

At x = 0.5 m, the field from the charge at x = 0:
${\vec{E}}_{2} = \frac{k q}{r} ^ 2 = \frac{k \cdot \left(- 4 \cdot {10}^{-} 6 Q\right)}{0.5 m} ^ 2$
${\vec{E}}_{2} = \frac{k \cdot 4 \cdot \left(- 4 \cdot {10}^{-} 6 Q\right)}{m} ^ 2 = \frac{k \cdot \left(- 16 \cdot {10}^{-} 6 Q\right)}{m} ^ 2$
This will also be to the left at x = 0.5 m.

So the sum is $\frac{k \cdot 71.5 \cdot {10}^{-} 6 Q}{m} ^ 2$ pointing left

b. At x = 1 m, the field from the charge at x = 0.8 m:
${\vec{E}}_{1} = \frac{k q}{r} ^ 2 = \frac{k \cdot 5 \cdot {10}^{-} 6 Q}{0.2 m} ^ 2$
${\vec{E}}_{1} = \frac{k \cdot 25 \cdot 5 \cdot {10}^{-} 6 Q}{m} ^ 2 = \frac{k \cdot 125 \cdot {10}^{-} 6 Q}{m} ^ 2$
This will be to the right at x = 1 m.

At x = 1 m, the field from the charge at x = 0:
${\vec{E}}_{2} = \frac{k q}{r} ^ 2 = \frac{k \cdot \left(- 4 \cdot {10}^{-} 6 Q\right)}{1 m} ^ 2$
${\vec{E}}_{2} = \frac{k \cdot 1 \cdot \left(- 4 \cdot {10}^{-} 6 Q\right)}{m} ^ 2 = \frac{k \cdot \left(- 4 \cdot {10}^{-} 6 Q\right)}{m} ^ 2$
This will be to the left at x = 1 m.

So the sum is $\frac{k \cdot 121 \cdot {10}^{-} 6 Q}{m} ^ 2$ pointing right.

I hope this helps,
Steve