Yer Madder said:Determine how far apart two parallel plates must be situated so that a potential difference of 2.50*10^2 V produces an electric field strength of 2.00*10^3 N/C
F=(k*Q1*Q2)/r^2
PD=kQ/r
E=kQ/r^2
I cannot even figure out which formulae to combine...
An electric field is a region around a charged object where other charged particles experience a force. The strength and direction of the electric field is determined by the magnitude and sign of the charges creating it.
Electric potential difference is the change in electric potential energy per unit charge as a charged particle moves from one point to another in an electric field. It is directly proportional to the strength of the electric field. The formula for calculating electric potential difference is V = Ed, where V is the potential difference, E is the electric field strength, and d is the distance between the two points.
Yes, the electric fields from multiple charged objects can be combined using vector addition. This means that the total electric field at a point is the sum of the individual electric fields from each charged object.
The electric field due to a point charge can be calculated using the formula E = kq/r^2, where E is the electric field strength, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), q is the charge of the object, and r is the distance between the point charge and the location where the electric field is being measured.
The relationship between electric potential and electric field is given by the equation E = -dV/dr, where E is the electric field strength, V is the electric potential, and r is the distance from the source of the electric field. This means that the electric field is the negative gradient of the electric potential. In other words, the electric field points in the direction of decreasing electric potential.