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A positive charge [tex]\( q_1 \)[/tex] of [tex]\( 5 \, \mu C \)[/tex] is [tex]\( 3 \times 10^{-2} \, m \)[/tex] west of a positive charge [tex]\( q_2 \)[/tex] of [tex]\( 2 \, \mu C \)[/tex].

What is the magnitude and direction of the electrical force [tex]\( F_e \)[/tex] applied by [tex]\( q_1 \)[/tex] on [tex]\( q_2 \)[/tex]?

A. Magnitude: [tex]\( 3 \, N \)[/tex], Direction: East
B. Magnitude: [tex]\( 3 \, N \)[/tex], Direction: West
C. Magnitude: [tex]\( 100 \, N \)[/tex], Direction: East
D. Magnitude: [tex]\( 100 \, N \)[/tex], Direction: West



Answer :

GinnyAnswer
Certainly! Let's solve the problem step by step.

We are given:

- [tex]\( q_1 = 5 \mu C = 5 \times 10^{-6} \)[/tex] C
- [tex]\( q_2 = 2 \mu C = 2 \times 10^{-6} \)[/tex] C
- The distance between the charges [tex]\( r = 3 \times 10^{-2} \)[/tex] m

We need to find the magnitude and direction of the electrical force [tex]\( F_e \)[/tex] applied by [tex]\( q_1 \)[/tex] on [tex]\( q_2 \)[/tex].

#### Step 1: Use Coulomb's Law
Coulomb's law states that the magnitude of the electrical force between two point charges is given by:
[tex]\[ F_e = k \frac{q_1 q_2}{r^2} \][/tex]
where [tex]\( k \)[/tex] is Coulomb's constant, approximately [tex]\( 8.99 \times 10^9 \)[/tex] N·m²/C².

#### Step 2: Plug in the values
Let's plug in the given values into Coulomb's law formula:
[tex]\[ q_1 = 5 \times 10^{-6} \text{ C} \][/tex]
[tex]\[ q_2 = 2 \times 10^{-6} \text{ C} \][/tex]
[tex]\[ r = 3 \times 10^{-2} \text{ m} \][/tex]
[tex]\[ k = 8.99 \times 10^9 \text{ N·m}^2/\text{C}^2 \][/tex]

[tex]\[ F_e = 8.99 \times 10^9 \times \frac{(5 \times 10^{-6})(2 \times 10^{-6})}{(3 \times 10^{-2})^2} \][/tex]

#### Step 3: Calculate the magnitude
By calculating the above expression, we get:
[tex]\[ F_e = 99.86168652631305 \text{ N} \][/tex]

So, the magnitude of the electrical force [tex]\( F_e \)[/tex] is approximately [tex]\( 100 \text{ N} \)[/tex].

#### Step 4: Determine the direction
Both charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are positive. Since like charges repel each other, the force exerted by [tex]\( q_1 \)[/tex] on [tex]\( q_2 \)[/tex] will be directed away from [tex]\( q_1 \)[/tex]. Given that [tex]\( q_1 \)[/tex] is west of [tex]\( q_2 \)[/tex], the force on [tex]\( q_2 \)[/tex] will be directed towards the east.

Thus, the magnitude and direction of the electrical force [tex]\( F_e \)[/tex] applied by [tex]\( q_1 \)[/tex] on [tex]\( q_2 \)[/tex] are:
[tex]\[ \text{Magnitude: } 100 \text{ N} \][/tex]
[tex]\[ \text{Direction: } \text{east} \][/tex]

So, the correct answer is:
[tex]\[ \text{magnitude: } 100 \text{ N} \][/tex]
[tex]\[ \text{direction: } east \][/tex]

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