Question

Is there a point between a 10 nc charge and a 20 nc charge at which the electric field is zero?

Answer 1

There exists a point between the two charges where the net electric field due to the charges is zero.

Explanation:

The electric field intensity due to a point charge is the number of electric field lines coming out of the charge and experienced at a point which is at a particular distance from the charge.

The two point charges having the same nature of charge will have the electric filed lines being repelled by one another. The intensity of the electric field will be greater due to the particle having higher charge as compared to the particle with lower charge.

It means that there will be a point near the charge with lower magnitude where the net electric field due to the two charges will be zero. The electric field lines due to the $20\text{ nC}$ will be more intensive and will dominate the region  between the two charges and the electric field lines of the charge $10\text{ nC}$ exist in smaller portion.

So, there will be a point near the $10\text{ nC}$ charge where the net electric field due to the charge will be zero.

Learn More:

1. The electric produced by the charge at point P brainly.com/question/11098511

2. Magnitude of charge on each plate of the capacitor brainly.com/question/8892837

3. Magnitude of electric field between two charges brainly.com/question/11220947

Answer Details:

Grade: High School

Subject: Physics

Chapter: Electrostatics

Keywords:

point, charge, 10 nC, 20 nC, electric field, intensity, portion, zero, net field, magnitude, nature, distance, field lines.

Answer 2

Let 10 nC and 20 nC charge is placed at "d" distance

now let say we will have a point at distance "x" from 10 nC at which electric field will be zero

now we will have

$E_1 = E_2$

$\frac{kq_1}{x^2} = \frac{kq_2}{(d-x)^2}$

now plug in all data

$\frac{k(10 nC)}{x^2} = \frac{k(20 nC)}{(d-x)^2}$

$\frac{10}{x^2} = \frac{20}{(d-x)^2}$

square root both sides

$\frac{1}{x} = \frac{\sqrt2}{d - x}$

$d - x = \sqrt2 x$

$d = x(1 + \sqrt2)$

$x = \frac{d}{1 + \sqrt2}$

so yes there will exist a point between 10 nC and 20 nC charge at which electric field will be zero