Question

Two charges +5μC and +20μC are placed 15 cm apart. At what point on the line joining the two charges is the electric field zero?

Answer 1

Answer:

5 +98 .309./as complete

Answer 2

Answer: There is no point on the line joining the charges where the electric field is zero.

Explanation:

To find the point on the line joining the two charges where the electric field is zero, we can use the principle of superposition. The electric field at any point on the line is the vector sum of the electric fields due to each charge separately.

Let's assume that the +5µC charge is at the origin (0, 0) and the +20µC charge is located at (0.15 m, 0), with the positive x-direction being to the right.

The electric field due to a point charge can be calculated using Coulomb's Law:

E = k * Q / r^2

Where:

E is the electric field,

k is the Coulomb's constant (9 x 10^9 N m^2/C^2),

Q is the charge, and

r is the distance between the point charge and the point where the electric field is being calculated.

Let's calculate the electric fields due to each charge separately at an arbitrary point on the x-axis (x, 0).

For the +5µC charge:

E1 = k * Q1 / r1^2

= (9 x 10^9 N m^2/C^2) * (5 x 10^(-6) C) / (x^2)

For the +20µC charge:

E2 = k * Q2 / r2^2

= (9 x 10^9 N m^2/C^2) * (20 x 10^(-6) C) / ((0.15 - x)^2)

To find the point where the electric field is zero, the vector sum of E1 and E2 should be zero:

E1 + E2 = 0

(9 x 10^9 N m^2/C^2) * (5 x 10^(-6) C) / (x^2) + (9 x 10^9 N m^2/C^2) * (20 x 10^(-6) C) / ((0.15 - x)^2) = 0

Simplifying the equation, we get:

5 / x^2 + 20 / (0.15 - x)^2 = 0

To solve this equation, we can multiply both sides by x^2 (0.15 - x)^2 to eliminate the denominators:

5 * (0.15 - x)^2 + 20 * x^2 = 0

Expanding and rearranging, we have:

0.75 - 1.5x + x^2 + 20x^2 = 0

21x^2 - 1.5x + 0.75 = 0

Now, we can solve this quadratic equation to find the values of x where the electric field is zero. We can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 21, b = -1.5, and c = 0.75.

x = (-(-1.5) ± √((-1.5)^2 - 4 * 21 * 0.75)) / (2 * 21)

x = (1.5 ± √(2.25 - 63)) / 42

x = (1.5 ± √(-60.75)) / 42

Since the discriminant (√(b^2 - 4ac)) is negative, the equation does not have any real solutions. This means there is no point on the line joining the charges where the electric field is exactly zero.