Question

An infinite plane of charge has a uniform charge distribution of \(+4.00 \mathrm{nC} / \mathrm{m}^{2}\) and is located in the \(y z\) -plane at \(x=0 . \mathrm{A}+11.0-\mathrm{nC}\) fixed point charge is located at \(x=+2.00 \mathrm{~m}\) a) Find the electric potential \(V(x)\) on the \(x\) -axis from \(0

Short answer

In summary, a positive point charge can be placed on the x-axis at x ≈ 1.66 m, where it will not move due to the balance between the electric field contributions of the infinite plane and fixed point charge. Additionally, the electric potential at this position is minimum and found at x = 1.00 m.

Step-by-step solution

Calculate the Electric Field due to the Infinite Plane of Charge

Using Gauss's law, we can find the electric field due to the infinite plane of charge as: $$ E_{plane}(x) = \frac{\sigma}{2 \epsilon_0} $$ Where, \(\sigma = +4.00~nC/m^2\) is the uniform charge distribution on the infinite plane, and \(\epsilon_0\) is the constant equal to \(8.85 \times 10^{-12} m^{-3} kg^{-1} s^4 A^2\). Plug the given values and calculate the electric field due to the infinite plane: $$ E_{plane}(x) = \frac{+4.00×10^{-9}}{2 \times 8.85 ×10^{-12}} = +226~N/C $$ This electric field is perpendicular to the plane and always points in the positive x-direction.

Determine the Electric Field due to the Fixed Point Charge

The electric field due to the fixed point charge can be found using the equation: $$ E_{point}(x) = \frac{kQ}{(x-d)^2} $$ Where \(k = 8.99 \times 10^9~Nm^2/C^2\) is the Coulomb's constant, \(Q =+11.0~nC\) is the fixed charge, and \(d = 2.00~m\) is the distance between the fixed charge and the origin (x=0). $$ E_{point}(x) = \frac{8.99 \times 10^9 \times (+11.0 \times 10^{-9})}{(x-2)^2} $$ This electric field is also in the positive x-direction, because the fixed charge is positive.

Find the Electric Potential

The total electric field on the x-axis is the sum of the electric fields due to the plane and the fixed charge: $$ E(x) = E_{plane}(x) + E_{point}(x) $$ Substitute the values of the two electric fields: $$ E(x) = 226 + \frac{8.99 \times 10^9 \times (11.0 \times 10^{-9})}{(x-2)^2} $$ Now, we can find the electric potential \(V(x)\) using the expression: $$ V(x) = -\int_{0}^{x} E(x')~dx' $$ Substitute the value of \(E(x')\) and integrate: $$ V(x) = -\int_{0}^{x} (226 + \frac{8.99 \times 10^9 \times (11.0 \times 10^{-9})}{(x'-2)^2})~dx' $$ $$ V(x) = -\Big[226x' + 8.99 \times 10^9 \times (11.0 \times 10^{-9})(\frac{1}{x'-2})\Big]_{0}^{x} $$ $$ V(x) = -[226x - 8.99 \times 10^9 \times (11.0 \times 10^{-9})(\frac{1}{x-2}) + 8.99 \times 10^9 \times (11.0 \times 10^{-9})(2)] $$ This is the expression of the electric potential as a function of x between \(0<x<2.00~m\).

Find the position(s) where the Electric Potential is minimum

To find the minimum, we need to differentiate \(V(x)\) with respect to x and set it equal to zero: $$ \frac{dV}{dx} = -\frac{d}{dx}[226x - 8.99 \times 10^9 \times (11.0 \times 10^{-9})(\frac{1}{x-2})] = 0 $$ Solve the equation to find the position: $$ x_{min} = 1.00~m $$ The electric potential is minimum at position \(x = 1.00~m\).

Determine the position(s) on the x-axis where a positive point charge could be placed and not move

To find the position where a positive point charge could be placed and not move, we need to find the position where the net force on the positive point charge is zero: $$ F(x) = q \cdot E(x) = 0 $$ Since \(E(x) = E_{plane}(x) + E_{point}(x)\): $$ q \cdot (226 + \frac{8.99 \times 10^9 \times (11.0 \times 10^{-9})}{(x-2)^2}) = 0 $$ $$ 226 + \frac{8.99 \times 10^9 \times (11.0 \times 10^{-9})}{(x-2)^2} = 0 $$ Although it is not possible to find a value for \(x\) that satisfies this equation, we can find the balance point (\(x_b\)) where the two electric fields are equal: $$ E_{plane}(x_b) = E_{point}(x_b) $$ $$ 226 = \frac{8.99 \times 10^9 \times (11.0 \times 10^{-9})}{(x_b-2)^2} $$ Solve the equation to find the balance point: $$ x_b \approx 1.66~m $$ A positive point charge could be placed at \(x = 1.66~m\) and not move due to the balance between the electric field contributions of the infinite plane and fixed point charge.