A charge of 10.104 nC is uniformly distributed along the x-axis from −1 m to 1 m.
What is the electric potential (relative to zero at infinity) of the point at 5 m on the x-axis?
Okay, so
V=[(8.98755e9)*(10.104e-9)*sqrt(5^2 + x^2)] from -1 to 1?
or
V=[(8.98755e9)*(10.104e-9)*sqrt(2^2 + x^2)] from 0 to 5?
where 2 is the distance from x=-1 to x=1
To determine the electric potential at a point due to a uniformly distributed charge, we need to integrate the contributions from individual charge elements along the x-axis. The formula for electric potential due to a point charge is:
V = k * (Q / r)
Where:
V is the electric potential
k is the electrostatic constant (8.99 × 10^9 Nm^2/C^2)
Q is the charge of the element
r is the distance between the charge element and the point at which we want to find the potential.
In this case, the charge is uniformly distributed along the x-axis. To find the electric potential at a point, we need to divide the charge into small elements and integrate their contributions.
Let's set up the integration. We divide the x-axis into many small charge elements Δx. The distance between a charge element and the point at 5 m on the x-axis is (5 - x). The charge of each element is given by ΔQ = q * Δx, where q is the charge per unit length:
q = Q / (2 * a)
Where:
Q = 10.104 nC (total charge)
a = 1 m (length of the charge distribution)
Now, we can integrate these charge elements to find the electric potential at the point.
V = ∫ [k * (q * Δx) / (5 - x)] from x = -1 to x = 1
To evaluate this integral, we need to substitute q and integrate from -1 to 1.
V = k * ∫ [((Q / (2 * a)) * Δx) / (5 - x)] from x = -1 to x = 1
Simplifying the integral further:
V = k * (Q / (2 * a)) * ∫ [(Δx) / (5 - x)] from x = -1 to x = 1
Now, we can evaluate this integral to get the electric potential at the point.