Question

The distance between two charges q1 = + 2 μC and q2 = + 6 μC is 15.0 cm. Calculate the distance from charge q1 to the points on the line segment joining the two charges where the electric field is zero.​

Answer 1

Answer:

The distance of the point is 5.5 cm from the charge q1.

Explanation:

Let a point is at the distance of x from the q1.

As the net electric field at the point is zero.

Thus, the electric field due to the charge q1 is equal and opposite to the electric field due to the charge q2.

$\frac{kq1}{ {(x)}^{2} } = \frac{kq2}{ {(d - x)}^{2} }$

where k is the electrostatic force constant and d is the distance between the charges.

Substituting the known values,

$\frac{q1}{ {(x)}^{2} } = \frac{q2}{ {(d - x)}^{2} } \\ \frac{2 \times {10}^{ - 6} }{{(x)}^{2} } = \frac{ 6\times {10}^{ - 6}}{ {(15 \times {10}^{ - 2} - x)}^{2} } \\ \frac{2 }{{(x)}^{2} } = \frac{ 6}{ {(0.15 - x)}^{2} } \\ 6 \times {(x)}^{2} = 2 \times {(0.15 - x)}^{2} \\ 3 \times {(x)}^{2} = {(0.15 - x)}^{2} \\ 1.732 \times x = 0.15 - x \\x(1.732 + 1) = 0.15 \\ 2.732 \times x = 0.15 \\ x = \frac{0.15}{2.732} \\ x = 0.055 m \\ x = 5.5cm$

Thus, the distance of point from the charge q1 is 5.5 cm.