© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Charges and Their Positions
• Charge A = q , positioned at x = 0 .
• Charge B = -2q , positioned at x = \frac{3}{4}R .
• Charge C = 2q , positioned at x = R .
Given data:
• q = 2 \times 10^{-6}\,\mathrm{C} ,
• R = 2\,\mathrm{cm} .
We want the net force on the charge -2q (charge B).
Step 2: Recall Coulomb’s Law
Coulomb’s law states that the electrostatic force F between two charges Q_1 and Q_2 , separated by a distance d , is:
F = \frac{K \, Q_1 \, Q_2}{d^2},
where K = 9 \times 10^9 \,\mathrm{N\,m^2\,C^{-2}} .
Step 3: Calculate the Force on -2q due to q
Denote this force by F_{BA} .
• Charge A = q at x=0 .
• Charge B = -2q at x=\frac{3}{4}R .
The distance between them is \frac{3}{4}R . Using the magnitude form of Coulomb’s law:
F_{BA}
= \left|\frac{K \, q \, (-2q)}{\left(\frac{3}{4}R\right)^2}\right|
= \frac{2K \, q^2}{\left(\frac{3}{4}R\right)^2}.
Since \left(\frac{3}{4}R\right)^2 = \frac{9}{16}R^2 , we get
F_{BA}
= \frac{2K \, q^2}{\frac{9}{16}R^2}
= \frac{32K \, q^2}{9R^2}.
This force acts to the left on -2q (they attract each other because one is positive and the other is negative).
Step 4: Calculate the Force on -2q due to 2q
Denote this force by F_{BC} .
• Charge B = -2q at x=\frac{3}{4}R .
• Charge C = 2q at x=R .
The distance between them is R - \frac{3}{4}R = \frac{1}{4}R . Thus,
F_{BC}
= \left|\frac{K \, (-2q) \, (2q)}{\left(\frac{1}{4}R\right)^2}\right|
= \frac{4K \, q^2}{\left(\frac{1}{4}R\right)^2}.
Since \left(\frac{1}{4}R\right)^2 = \frac{1}{16}R^2, we have
F_{BC}
= \frac{4K \, q^2}{\frac{1}{16}R^2}
= 64\frac{K \, q^2}{R^2}.
This force acts to the right on -2q (again attractive).
Step 5: Determine the Net Force on -2q
Since F_{BA} is to the left and F_{BC} is to the right, the net force on -2q is:
F_{B} = F_{BC} - F_{BA}.
Substitute the expressions:
F_{B} = \frac{64K \, q^2}{R^2} - \frac{32K \, q^2}{9R^2}
= \frac{64K \, q^2 \times 9 - 32K \, q^2}{9R^2}.
That is:
F_{B} = \frac{576K \, q^2 - 32K \, q^2}{9R^2}
= \frac{544K \, q^2}{9R^2}.
Step 6: Substitute Numerical Values
• K = 9 \times 10^9 \,\mathrm{N\,m^2\,C^{-2}}
• q = 2 \times 10^{-6}\,\mathrm{C}
• R = 2\,\mathrm{cm} = 2 \times 10^{-2}\,\mathrm{m}
So,
F_{B}
= \frac{544 \times 9 \times 10^9 \times (2 \times 10^{-6})^2}{9 \times (2 \times 10^{-2})^2}.
1. Compute (2 \times 10^{-6})^2 = 4 \times 10^{-12} .
2. Compute (2 \times 10^{-2})^2 = 4 \times 10^{-4} .
3. Numerator = 544 \times 9 \times 10^9 \times 4 \times 10^{-12} = 544 \times 36 \times 10^{-3} = 19.584 .
4. Denominator = 9 \times 4 \times 10^{-4} = 36 \times 10^{-4} = 3.6 \times 10^{-3} .
Hence,
F_{B}
= \frac{19.584}{3.6 \times 10^{-3}}
\approx 5440\,\mathrm{N}.
Final Answer
The magnitude of the net force on the charge -2q is approximately 5440\,\mathrm{N} (towards the right).
