An equipotential surface that surrounds a point charge q has a potential of 590 V and an area of 1.20 m2. Determine q.
A= 4πr² =>
r=sqrt(A/4π )= sqrt(1.2/4•3.14)= 0.31 m
φ=q/4πε₀r =>
q= 4πε₀rφ=
=4π•8.85•10⁻¹²•0.31•590=2.03•10⁻⁸ C
I’m stupid
To solve this problem, we need to recall that the potential (V) of a point charge is given by the equation:
V = k * (q / r)
where V is the potential, k is the electrostatic constant (9 × 10^9 N·m^2/C^2), q is the charge of the point charge, and r is the distance from the point charge.
In this case, we are given the potential V and the area A of the equipotential surface. However, we need more information to determine the charge q.
The equation relating potential, area, and charge for an equipotential surface is:
V = k * (q / A)
where V is the potential, k is the electrostatic constant, q is the charge on the surface, and A is the area of the surface.
We can rearrange this equation to solve for q:
q = V * (A / k)
Now, substituting the given values:
q = 590 V * (1.20 m^2 / (9 × 10^9 N·m^2/C^2))
Calculating this expression yields the value of q.