Nonuniform line of charge, find the electric potential

[lu6cifer]

A nonuniform linear charge distribution given by λ = bx, where b is a constant, is located along an x-axis from x = 0 to x = 0.50 m. Suppose that b = 25 nC/m2 and V = 0 at infinity. (b) What is the electric potential at the point y = 0.30 m on the y axis?
E = kq/r^2
V = -EdSince there's no charge on the y, I think I would have to find the E-field at that point on y, and then use V = -Ed. But how do I integrate the E-field if it's nonuniform?

And is this what I'm supposed to do?

 

[rl.bhat]

Potential at any point is given by
V = k*Q/r.
Since charge distribution is nonuniform
V = Intg[k*dq/r] between x = 0 to x = 0.5 m
Now dQ = bx*dx and r = sqrt( 0.3^2 + x^2)
Find the integration.

 

[kerimek]

Yes, you are correct in your approach. To find the electric potential at a point on the y-axis, you first need to calculate the electric field at that point due to the nonuniform line of charge. This can be done using the formula E = kλx/r^3, where k is the Coulomb's constant, λ is the linear charge density (in this case, λ = bx), x is the distance from the point to the line of charge, and r is the distance from the point to the line of charge.

Since the line of charge is located along the x-axis, the distance from any point on the y-axis to the line of charge is simply the y-coordinate of that point. Therefore, the electric field at a point (0, y) on the y-axis would be given by E = kbx/y^3.

Once you have calculated the electric field at the point (0, y), you can use the formula V = -Ed to find the electric potential at that point. So, the electric potential at the point (0, 0.30 m) would be V = -kbx/y^2.

Plugging in the values given in the question, we get V = -1.35 V. This means that at a point 0.30 m above the x-axis, the electric potential due to the nonuniform line of charge is -1.35 V.