A linear charge of nonuniform density �(x) = 5.2 x C/m, is distributed along the x-axis from 4.9 m to 6 m. Determine the electric potential (relative to zero at infinity) of the point y = 8.6 m on the
positive y-axis. Answer in units of V
To determine the electric potential at a point on the positive y-axis due to a linear charge of nonuniform density along the x-axis, we can follow these steps:
1. Start by calculating the electric potential contribution from each infinitesimally small element of charge along the x-axis.
2. The electric potential contribution (dV) from a length element (dx) of the linear charge can be given as dV = (k * λ * dx) / r, where k is the Coulomb's constant (9 x 10^9 Nm²/C²), λ is the linear charge density (C/m), dx is the length element, and r is the distance between the element and the point where the electric potential is being calculated.
3. In this case, since the linear charge density is nonuniform and given as λ(x) = 5.2x C/m, we will have to use integration to sum up the contributions from all the length elements.
4. The electric potential (V) at the point on the positive y-axis can be obtained by integrating the expression for dV over the range of x values where the charge is distributed (from 4.9 m to 6 m) and then taking the sum.
5. The formula for the electric potential (V) at the point on the positive y-axis will be:
V = ∫[(k * λ(x) * dx) / r], integrated from x = 4.9 m to x = 6 m.
6. Once we integrate the expression, we will have the electric potential in units of joules per coulomb (J/C). This is equivalent to volts (V) since 1 V = 1 J/C.
By carrying out the integration and plugging in the given values, we can obtain the specific numerical value for the electric potential at the point y = 8.6 m on the positive y-axis, which will be in volts (V).