A linear charge of nonuniform density �lambda(x)=bx C/m, where b = 5.2 nC/m2, 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.
Electric potential is scalar, so you just add the incremetnal parts.
V= k INT dq/distance
ok, distance= sqrt(y^2+x^2)
and dq=lambda*x*dx
V=k int lamda x dx /sqrt(y^2+x^2)
V=k*lambda INT x/( ) dx
= k lambda sqrt ( ) over the limits
check all that.
So going on that
V=[(8.98755e9)*(5.2e-9)*sqrt(8.6^2 + x^2)] from 4.9 to 6?
:)))))))))))))))))))))))))))GAU
To determine the electric potential at a point due to a linear charge with non-uniform density, we can use the principle of superposition. This principle states that the total potential at a point due to multiple charges is the sum of the individual potentials due to each charge.
In this case, we need to calculate the potential at a point y = 8.6 m on the positive y-axis due to the linear charge distribution along the x-axis from 4.9 m to 6 m.
To start, let's break down the linear charge distribution into small elemental charges. We can consider each small segment of the charge distribution as a point charge and calculate the potential due to each segment.
The charge density lambda(x) is given as bx C/m, where b = 5.2 nC/m². The elemental charge dq at a distance x is given by dq = lambda(x) dx.
Now, let's calculate the potential at the point y = 8.6 m due to each elemental charge.
The potential at a point due to a point charge q at a distance r from the point charge is given by V = k * q / r, where k is the electrostatic constant (k = 9 × 10^9 Nm²/C²).
We can integrate the potential due to each elemental charge along the x-axis from 4.9 m to 6 m to get the total potential at the point y = 8.6 m.
∫ V(x) dx = k * ∫ (dq / r)
Since we are calculating the potential at a point on the y-axis, the distance r between the elemental charge and the point is the perpendicular distance from the elemental charge to the point on the y-axis. In this case, the distance r is the difference between y and x.
∫ V(x) dx = k * ∫ (lambda(x) dx / (y - x))
Substituting the given charge density lambda(x) = bx, we have:
∫ V(x) dx = k * ∫ (bx dx / (y - x))
Now, we can plug in the values and calculate the potential at the point y = 8.6 m.
y = 8.6 m
b = 5.2 nC/m²
k = 9 × 10^9 Nm²/C²
r = y - x
∫ V(x) dx = k * ∫ (bx dx / (y - x)) from x = 4.9 m to 6 m
Evaluate the integral to find the potential at the point y = 8.6 m.
The result of the integral will give us the electric potential (relative to zero at infinity) of the point y = 8.6 m on the positive y-axis.