Question

Two charges lie on the x axis, +3q at the origin, and -2q at x = 5.0 m. The point on the x axis where the electric potential has a zero value (when the value at infinity is also zero) is
a. 1.0 m.b. 2.0 m.c. 2.5 m.d. 3.0 m.e. 4.0 m.

Answer 1

Answer:

Explanation:

q1 = 3q at (0,0)

q2 = - 2q at (5, 0)

Let the potential is zero at a distance x from the origin

$V_{1}+V_{2}=0$

$\frac{k\times 3q}{x}-\frac{k\times 2q}{5-x}=0$

15 - 3x = 2x

5x = 15

x  = 3 m

Thus, the potential is zero at x = 3 m .

Option (d) is true.

Answer 2

The point on the x-axis should be option d 3.0m.

What is Electric Field:

It is the property of charged particles that represent the attraction or repulsion of charges. Since the electric field should be considered like the vector, the resultant electric field because of the configuration of charges would be equivalent to the vector sum of all individual electric fields of the charges.

Calculation of the point:

Since

q1 = 3q at (0,0)

q2 = - 2q at (5, 0)

Now

Let us assume the potential is zero at a distance x from the origin

so,

$V_1 + V_2 = 0\\\\\frac{k\times3q}{x} - \frac{k\times2q}{5-x} = 0$

Now

15-3x = 2x

15 = 2x + 3x

15 = 5x

x = 3m

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