Question

An electric point charge of Q1 = 7.43 nC is placed at the origin of the real axis. Another point charge of Q2 = 1.21 nC is placed at a position of p = 3.95 m on the real axis. At which position can a third point charge of q = -1.99 nC be placed so that the net electrostatic force on it is zero?

Answer 1

Answer:

attached below

Answer 2

The third point charge of q = -1.99 nC should be placed at a distance of 2.8m on the real axis so that the net electrostatic force on it is zero.

According to the question, the given charges their positions are:

Q₁ = 7.43 nC = 7.43 × 10⁻⁹ C  is at the origin, r₁ = 0m

Q₂ = 1.21 nC = 1.21 × 10⁻⁹ C is at r₂ = 3.95m

The test q = -1.99 nC = 1.99 × 10⁻⁹ C.

Since the test charge q is negatively charged, and both the charges Q₁ and Q₂ are positively charged, they both will attract the charge q.

Therefore to have a zero net electrostatic force on the charge q it must be placed between the two charges Q₁ and Q₂ as shown in the image attached.

Let the charge q is placed at a distance x from the origin.

The force on charge q due to charge Q₁:

$F_1=k\frac{qQ_1}{x^2}\\ \\ F_1 = \frac{9*10^9*1.99*7.43*10^{-18}}{x^2}\\\\ F_1 = \frac{13.3*10^{-8}}{x^2}N$

The force on charge q due to charge Q₂:

$F_2=k\frac{qQ_2}{(3.95-x)^2}\\ \\F_2 = \frac{9*10^9*1.99*1.21*10^{-18}}{(3.95-x)^2}\\\\F_2 = \frac{2.17*10^{-8}}{(3.95-x)^2}N$

for the charge q to be in equilibrium:

F₁ = F₂

$\frac{6}{x^2}=\frac{1}{(3.95-x)^2} \\ \\ (3.95-x)^2*6-x^2=0$

solving the quadratic equation we get:

$x = 2.8m\\ \\ or\\ \\ x=6.67m$

but  x = 6.67m is will be rejected because at this point both the forces will be in same direction.

So x = 2.8m should be the position of charge q.

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