Question

two unlike charges separated by a distance of 1 meter attract eachother witha a force .108N if the charges are in ratio 1:3 the weak chage is

Answer 1

2*10^-6

Answer 2

The problem involves calculating the magnitude of two unlike charges given their ratio, the distance separating them, and the force of attraction between them. We will use Coulomb's Law for this calculation.

1. Identify Given Information:

• Nature of charges: Unlike (one positive, one negative).
• Distance ($r$) = 1 meter.
• Force of attraction ($F$) = 0.108 N.
• Ratio of magnitudes of charges ($|q_1| : |q_2|$) = 1:3.

2. Define Variables: Let the magnitudes of the two charges be $|q_1|$ and $|q_2|$. From the ratio, we can write $|q_1| = x$ and $|q_2| = 3x$, where $x$ is a constant. The weaker charge is $x$.

3. Apply Coulomb's Law: Coulomb's Law states that the force ($F$) between two point charges is given by: $F = k \frac{|q_1 q_2|}{r^2}$ where $k$ is Coulomb's constant, $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.

4. Substitute Values and Solve for x: Substitute the given values into Coulomb's Law: $0.108 = (9 \times 10^9) \frac{(x)(3x)}{(1)^2}$ $0.108 = (9 \times 10^9) (3x^2)$ $0.108 = 27 \times 10^9 x^2$

Now, solve for $x^2$: $x^2 = \frac{0.108}{27 \times 10^9}$ To simplify the calculation, convert 0.108 to $108 \times 10^{-3}$: $x^2 = \frac{108 \times 10^{-3}}{27 \times 10^9}$ $x^2 = \left(\frac{108}{27}\right) \times 10^{-3} \times 10^{-9}$ $x^2 = 4 \times 10^{-12}$

Now, take the square root to find $x$: $x = \sqrt{4 \times 10^{-12}}$ $x = 2 \times 10^{-6} \text{ C}$

5. Determine the Weaker Charge: The weaker charge is $x$. Therefore, the weaker charge is $2 \times 10^{-6} \text{ C}$, which can also be written as $2 \mu\text{C}$ (microcoulombs).

The stronger charge would be $3x = 3 \times (2 \times 10^{-6} \text{ C}) = 6 \times 10^{-6} \text{ C} = 6 \mu\text{C}$.

The charges are unlike, so one charge is $+2 \mu\text{C}$ and the other is $-6 \mu\text{C}$ (or vice versa).