A positive point charge q = +2.50 nC is located at x = 1.20 m and a negative charge of ƒ{2q = ƒ{5.00 nC is located at the origin. (a) Sketch the electric potential verses x for points on the x-axis in the range -1.50 m < x < 1.50 m. (b) Find a symbolic expression for the potential on the x-axis at an arbitrary point P between the two charges. The
expression should be in terms of x, q and d = 1.20 m, the distance between the two charges. (c) Find the electric potential at x = 0.600 m, i.e. at x = d/2. (d) Find the value of x at the point along the x-axis between the two charges where the electric potential is zero.
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(a) To sketch the electric potential versus x for points on the x-axis, we need to use the equation for electric potential due to a point charge:
V = k * q / r
where V is the electric potential, k is the electrostatic constant (k = 9 * 10^9 N m^2 / C^2), q is the charge magnitude, and r is the distance from the charge.
For positive charge q = +2.50 nC located at x = 1.20 m, the electric potential at any point on the x-axis is given by V1 = k * (2.50 * 10^-9 C) / r1, where r1 is the distance between the point on the x-axis and the positive charge (x value of the point).
For negative charge q = -5.00 nC located at the origin, the electric potential at any point on the x-axis is given by V2 = k * (-5.00 * 10^-9 C) / r2, where r2 is the distance between the point on the x-axis and the negative charge (x value of the point).
To sketch the electric potential versus x, we can choose some values of x within the range -1.50 m < x < 1.50 m and calculate the corresponding electric potentials V1 and V2 using the equations above. We can then plot the values on a graph, with x on the x-axis and V on the y-axis.
(b) To find a symbolic expression for the potential on the x-axis at an arbitrary point P between the two charges, we can use the principle of superposition. The total electric potential at point P would be the sum of the potentials due to the positive and negative charges:
Vp = V1 + V2
Substituting the equations for V1 and V2 from part (a), we have:
Vp = k * (2.50 * 10^-9 C) / r1 - k * (5.00 * 10^-9 C) / r2
Since P is an arbitrary point on the x-axis between the two charges, the distances r1 and r2 can be expressed in terms of x and d:
r1 = |x - 1.20 m|
r2 = |x|
Substituting these values in the expression for Vp, we get the symbolic expression for the potential on the x-axis at an arbitrary point P between the two charges.
(c) To find the electric potential at x = 0.600 m (x = d/2), we can substitute this value into the expression for Vp derived in part (b) and calculate it using the given charge magnitudes and distances.
(d) To find the value of x where the electric potential is zero on the x-axis between the two charges, we can set the expression for Vp derived in part (b) equal to zero and solve for x.