Electric Fields ☃️ Flashcards
Define electric field strength
Force per unit positive charge
F.S.
Ratio of resitance of filament to that of connecting wire is 1000.
Explain qualitatively why the power dissipated in the filament wire of the lamp is greater than the total power dissipated in the connecting wires. Both wires are connected in series
- Resistance of the filament is greater.
- Same current in both types of wires.
P = I² R
:. P > for filament
Note:
1. Equations given in part 1 can be used further down
2. Ratios given in part 1 can be used further down. So long as you have the ratio and at least one value, the other can be determined
One connecting wire and one filament eire connected to power supply and a lamp in series.
The diameter of the connecting wires is decreased. The total length of the connecting wires and the resistivity of the metal of the connecting wires remain the same.
State and explain the change, if any, that occurs to the resistance of the filament wire of the lamp.
R (connecting) ⬆️ as A ⬇️
I decreases
P.d decreases
:. R (filament) decreases
Define electric field
Region where a force acts on a stationary charge
Coulomb’s Law
The electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of their seperation
F = (1/4πε.) × Qq / r²
ε. = 8.85 × 10^-12
Electric field strength
E is the force experienced per unit positive charge placed at that pt.
E = F/q = (1/4πε.)× Q/r²
Q - charge causing the field
Unit: newton per coulomb
Electric potential energy
Electric potential energy of a charged body a point in an electric fi work done on the body in moving it from infinity to that point in the field.
Ep = qV
In general,
ΔU = qΔV
Where ΔU - change in energy
Electric potential
Work done per unit positive charge brought from infinity to a point in the electric field.
V = (1/4πε.)× Q/r
