Two fixed point charges are on the \(x\) -axis. A charge of \(-3.00 \mathrm{mC}\) is located at \(x=+2.00 \mathrm{~m}\) and a charge of \(+5.00 \mathrm{mC}\) is located at \(x=-4.00 \mathrm{~m}\) a) Find the electric potential, \(V(x),\) for an arbitrary point on the \(x\) -axis. b) At what position(s) on the \(x\) -axis is \(V(x)=0 ?\) c) Find \(E(x)\) for an arbitrary point on the \(x\) -axis.

To find the electric potential at any point on the x-axis, we first need to find the electric potential due to each point charge. For an arbitrary point x along the x-axis, the distance from the first charge to x is \(x - 2\) meters and the distance from the second charge to x is \(x + 4\) meters. The electric potential, V(x), due to a point charge q at a distance r, is given by: \(V(x) = \dfrac{kq}{r}\), where \(k = 8.99 * 10^9 N m^2/C^2\) is the Coulomb's constant. Thus, the electric potential due to each charge at an arbitrary point x can be found using the above formula: \(V_1(x) = \dfrac{k(-3.00 * 10^{-3} C)}{x - 2}\) (due to the first charge) \(V_2(x) = \dfrac{k(5.00 * 10^{-3} C)}{x + 4}\) (due to the second charge)

Now, to obtain the total electric potential (V(x)) at an arbitrary point x along the x-axis, we simply add the individual electric potentials due to each charge: \(V(x) = V_1(x) + V_2(x)\)

To find the position(s) along the x-axis where the electric potential is 0, we place the total electric potential V(x) equal to 0 and solve for x: \(0 = V_1(x) + V_2(x)\) Rearrange the equation: \(V_1(x) = -V_2(x)\) Then, solve for x to find the position(s) where the electric potential is 0.

To find the electric field (E(x)) at any point x along the x-axis due to these point charges, we use the equation for the electric field due to a point charge: \(E(x) = \dfrac{kq}{r^2}\) Hence, the electric field due to each charge at an arbitrary point x is given by: \(E_1(x) = - \dfrac{k(-3.00 * 10^{-3} C)}{(x - 2)^2}\) (due to the first charge) \(E_2(x) = \dfrac{k(5.00 * 10^{-3} C)}{(x + 4)^2}\) (due to the second charge)

Now to calculate the total electric field (E(x)) at an arbitrary point x along the x-axis, we must add the individual electric fields due to each charge. However, since we are only concerned with the x-component of the electric field for this problem, we sum the x-components of the electric fields: \(E(x) = E_1(x) + E_2(x)\)