# Question #9a373

Apr 30, 2017

$5.4 \cdot {10}^{12} N {C}^{-} 1$

#### Explanation:

We know that the force exerted on the test charge is
$\vec{F} = \vec{E} \cdot q$ (q is test charge)

here we got two charges applying the force on it in one direction . so net force on it is given by ,

$2 \cdot \vec{F} = \vec{E} \cdot q$
$\vec{E} = 2 \cdot \frac{\vec{F}}{q} = 2 \cdot k \cdot \frac{3 \cdot q}{0.10} ^ 2 \cdot \frac{1}{q} = \frac{9 \cdot {10}^{9} \cdot 1.5}{{10}^{-} 2}$

$5.4 \cdot {10}^{12} N {C}^{-} 1$

May 14, 2017

Showing that electric field is doubled when you have two charges:

#### Explanation:

To prove my point that electric field is doubled when there are two charges, I have used PHET SIMS. You can play with the simulation here:

In my first picture, I have placed one positive charge to the left. I have tried my best to center the positive charge on the arrow and the sensor three arrows away from the positive charge. I then measured the length of the arrow. The arrow represents the electric field strength.

In my second picture, I kept the positive charge and sensor at the same spot. I added a negative charge 3 arrows away from the sensor. If the electric field is not doubled, I shouldn't have to move the measuring tape. However, when I added the negative charge, I had to move the measuring tape. :

In our first picture, the length of the arrow was $44.8$ cm. In our second picture, the length of the arrow was $88.4$. This is roughly double the length.

The length is not exactly doubled because I may not have centered the charge. I have two uncertainties: I may have placed the negative charge at exactly the same distance as the positive charge and that my measuring with the measuring tape may have been off. However, it still suggests that electric field is doubled when there's two charges instead of one.

May 15, 2017

We know that Electric field $\vec{E}$ is defined as the electric force experienced by unit positive test charge.

The direction of the field is taken to be the direction of the electric force. Also that electric field is radially outward from a positive point charge and radially inward to a negative point charge.

The electric field due a number of point charges is obtained from vector sum of all individual fields. Sign convention followed is : A positive number is taken as an outward field; the field of a negative charge is directed towards it.

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Let charge $3 C$ be located at end $A$ and $- 3 C$ be located at the other end $B$ of a straight line $A B = 20 c m$ along $x$-axis. Let $A$ be origin of coordinates

Suppose $O$ is mid-point, so that $A O = O B = 10 c m$

From Coulombs law we have electric field at a distance $r$ from the point charge $Q$ as

$\vec{E} = \frac{Q}{4 \pi {\epsilon}_{0} {r}^{2}} \hat{r}$
where ${\epsilon}_{0}$= Permittivity of free space and $\frac{1}{4 \pi {\epsilon}_{0}} = 9 \times {10}^{9} N {m}^{2} {C}^{-} 2$

Inserting given values we get
1. Electric field at $O$ due to point charge at $A$
${\vec{E}}_{A \to O} = \frac{3 \times 9 \times {10}^{9}}{0.1} ^ 2 \hat{i}$
2. Similarly Electric field at $O$ due to point charge at $B$
${\vec{E}}_{B \to O} = \frac{3 \times 9 \times {10}^{9}}{0.1} ^ 2 \hat{i}$

Notice that $- v e$ sign of charge is absorbed in the direction of unit vector $\hat{i}$ due to sign convention.

Combined Electric Field at $O$
${\vec{E}}_{T} = \frac{3 \times 9 \times {10}^{9}}{0.1} ^ 2 \hat{i} + \frac{3 \times 9 \times {10}^{9}}{0.1} ^ 2 \hat{i}$
$\implies {\vec{E}}_{T} = \frac{2 \times 3 \times 9 \times {10}^{9}}{0.1} ^ 2 \hat{i}$
$\implies {\vec{E}}_{T} = 5.4 \times {10}^{12} \hat{i} V {m}^{-} 1$