Question

Two charges
Q1=500μC
and
Q2=100μC
are located on the XY plane at the positions
r1=3j⃗
m and
r2=4i⃗
m . Find the force exerted on the
Q

Answer 1

Answer on Question 58944, Physics, Electromagnetism

Question:

Two charges $Q_{1} = 500\mu C$ and $Q_{2} = 100\mu C$ are located on the XY plane at the positions $\vec{r_1} = 3\vec{j} m$ and $\vec{r_2} = 4\vec{i} m$ . Find the force exerted on the $Q_{2}$ .

Solution:

We can find the force exerted on the charge $Q_{2}$ from the Coulomb's law. Coulomb's law states that the force of attraction or repulsion between two electrically charged particles is directly proportional to the magnitude of their charges and inversely proportional to the square of the distance between them. Let's write the Coulomb's law in vector notation:

$$\overrightarrow {F _ {2 1}} = k \frac {Q _ {1} Q _ {2}}{r _ {2 1} ^ {2}} \overrightarrow {r _ {2 1}},$$

here, $\overrightarrow{F_{21}}$ is the force exerted on the charge $Q_{2}$ due to charge $Q_{1}$ , $k = 9\cdot 10^{9}N\frac{m^{2}}{C^{2}}$ is the Coulomb's constant, $Q_{1}$ , $Q_{2}$ is the charges, $\overrightarrow{r_{21}}$ is the unit vector, $r_{12}$ is the distance between two charges.

We defining the unit vector as follows:

$$\overrightarrow {r _ {2 1}} = \frac {\overrightarrow {r _ {2 1}}}{| \overrightarrow {r _ {2 1}} |} = \frac {\overrightarrow {r _ {2 1}}}{r _ {2 1}},$$

here, $|\overrightarrow{r_{21}}|$ is the magnitude of the $\overrightarrow{r_{21}}$ .

So, we can rewrite our formula:

$$\overrightarrow {F _ {2 1}} = k \frac {Q _ {1} Q _ {2}}{r _ {2 1} ^ {2}} \frac {\overrightarrow {r _ {2 1}}}{r _ {2 1}},$$

here, $\overrightarrow{r_{21}} = \overrightarrow{r_2} -\overrightarrow{r_1} = 4\vec{i} -3\vec{j}$ is the vectorial distance between two charges.

We can find the distance between two charges from the Pythagorean theorem:

$$r _ {2 1} = \sqrt {r _ {1} ^ {2} + r _ {2} ^ {2}} = \sqrt {(3 m) ^ {2} + (4 m) ^ {2}} = 5 m.$$

Substituting into the previous equation $r_{21}$ we get:

$$\begin{array}{l} \overrightarrow {F _ {2 1}} = k \frac {Q _ {1} Q _ {2}}{r _ {2 1} ^ {2}} \frac {\overrightarrow {r _ {2 1}}}{r _ {2 1}} = 9 \cdot 1 0 ^ {9} N \frac {m ^ {2}}{C ^ {2}} \cdot \frac {5 0 0 \cdot 1 0 ^ {- 6} C \cdot 1 0 0 \cdot 1 0 ^ {- 6} C}{(5 m) ^ {3}} \cdot (4 \vec {i} - 3 \vec {j}) = \\ = 3. 6 \cdot (4 \vec {i} - 3 \vec {j}) = (1 4. 4 \vec {i} - 1 0. 8 \vec {j}) N. \\ \end{array}$$

Also, we can find the magnitude of the force from the Pythagorean theorem:

$$F _ {2 1} = \sqrt {F _ {2 1 x} ^ {2} + F _ {2 1 y} ^ {2}} = \sqrt {(1 4 . 4 N) ^ {2} + (- 1 0 . 8 N) ^ {2}} = 1 8 N.$$

Answer:

$$\overrightarrow {F _ {2 1}} = (1 4. 4 \vec {i} - 1 0. 8 \vec {j}) N.$$

The magnitude of the force is $F_{21} = 18N$ .

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