Question

The distance between two charges \( q_1 = +2 \, \mu C \) and \( q_2 = +8 \, \mu C \) is 15 cm. Calculate the distance from the charge \( q_1 \) to the points on the line segment joining the two charges where the electric field is zero.

• A. 1 cm
• B. 2 cm
• C. 3 cm
• D. 4 cm
• E. 5 cm

Hint:

To find the point where the electric field is zero between two charges, set up an equation equating the electric fields due to both charges and solve for the distance.

Answer 1

Answer: (E)

Step 1: Concept of electric field due to point charges.}
The electric field at any point due to a point charge \( q \) is given by: \[ E = \frac{k |q|}{r^2} \] where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge to the point where the field is being calculated.
Step 2: Condition for zero electric field.}
For the electric field to be zero at some point between two charges, the electric fields due to both charges must be equal in magnitude and opposite in direction. Let the distance from charge \( q_1 \) to the point where the electric field is zero be \( r \). The distance from charge \( q_2 \) to the same point is \( (15 - r) \) cm, since the total distance between the charges is 15 cm.
Step 3: Setting up the equation.}
The magnitudes of the electric fields due to the charges must satisfy: \[ \frac{k |q_1|}{r^2} = \frac{k |q_2|}{(15 - r)^2} \] Substitute the values \( q_1 = 2 \, \mu C \) and \( q_2 = 8 \, \mu C \): \[ \frac{2}{r^2} = \frac{8}{(15 - r)^2} \] Cross-multiply to solve for \( r \): \[ 2(15 - r)^2 = 8r^2 \] Simplifying: \[ (15 - r)^2 = 4r^2 \] Expanding both sides: \[ 225 - 30r + r^2 = 4r^2 \] \[ 225 - 30r = 3r^2 \] \[ 3r^2 + 30r - 225 = 0 \] Divide the equation by 3: \[ r^2 + 10r - 75 = 0 \] Solve this quadratic equation using the quadratic formula: \[ r = \frac{-10 \pm \sqrt{10^2 - 4(A)(-75)}}{2(A)} = \frac{-10 \pm \sqrt{100 + 300}}{2} = \frac{-10 \pm \sqrt{400}}{2} \] \[ r = \frac{-10 \pm 20}{2} \] Thus, \( r = 5 \) or \( r = -15 \). Since distance cannot be negative, we have \( r = 5 \) cm.
Step 4: Final answer.}
Therefore, the distance from \( q_1 \) to the point where the electric field is zero is 5 cm from \( q_1 \).