Respuesta :

Answer:

Potential will be zero at two points

x = 3 cm

x = 6 cm

Explanation:

Let the first point at which potential is zero is lying between two charges

so we will have

[tex]\frac{kq_1}{x} = \frac{kq_2}{L - x}[/tex]

[tex]\frac{3nC}{x} = \frac{1nC}{4 - x}[/tex]

[tex]3(4 - x) = x[/tex]

[tex]x = 3 cm[/tex]

Let another point lies on the right side of -1 nC on x axis

so we will have

[tex]\frac{kq_1}{x} = \frac{kq_2}{x-4}[/tex]

[tex]\frac{3}{x} = \frac{1}{x-4}[/tex]

[tex]3(x - 4) = x[/tex]

[tex]x = 6 cm[/tex]

onyebuchinnaji

The points on x - axis in which the electric potential between the two charges is zero are 3 cm and 6 cm.

The given parameters;

q₁ = 3 nC

q₂ = -1 nC

Let the point in which the potential between the two charges equal zero, lie  between the two charges.

Let the point = x₁

         (+q₁)----------------------(x₁)------------------------(-q₂)

The electric potential of due to each charge is calculated as;

[tex]\frac{kq_1}{x_1} = \frac{kq_2}{4-x_1}[/tex]

[tex]\frac{3k}{x_1} = \frac{k}{4-x_1}\\\\3k(4-x_1) = kx_1\\\\3(4-x_1) = x_1\\\\12 - 3x_1 = x_1\\\\4x_1 = 12\\\\x_1 = \frac{12}{4} \\\\x_1 = 3 \ cm[/tex]

Since the second charge is negative, another point in which the potential between the two charges will be zero will be right of second charge;

          (+q₁)----------------------------(-q₂)------------------(x₂)

[tex]\frac{kq_1}{x_2} = \frac{kq_2}{x_2 - 4} \\\\\frac{3k}{x_2} = \frac{k}{x_2 - 4} \\\\3(x_2-4) = x_2\\\\3x_2 - 12 = x_2\\\\2x_2 = 12\\\\x_2 = \frac{12}{2} \\\\x_2 = 6 \ cm[/tex]

Thus, the points on x - axis in which the electric potential between the two charges is zero are 3 cm and 6 cm.

Learn more here: https://brainly.com/question/14427507

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