Question

Inkjet printers, the kind that most of us use, actually steer tiny drops of ink using electric fields! The drops of ink are given tiny electric charges and are propelled between metal plates 1.08 mm long, separated by a gap of 0.230 mm. If the electric field strength in the gap equals 2,248,335 V/m, what potential, V, in volts, must the printer circuits apply to the plates

Answer 1

To find the potential that the printer circuits must apply to the plates, we can use the formula for electric field strength (E):

E = V/d

where:
E is the electric field strength in volts per meter (V/m),
V is the potential in volts (V), and
d is the distance between the plates in meters (m).

In this case, the given electric field strength is 2,248,335 V/m, and the distance between the plates is 0.230 mm, which is equivalent to 0.230 × 10^(-3) meters.

Substituting these values into the formula, we can solve for V:

2,248,335 V/m = V / (0.230 × 10^(-3) m)

V = 2,248,335 V/m * 0.230 × 10^(-3) m

V ≈ 517.02 volts

Therefore, the printer circuits must apply a potential of approximately 517.02 volts to the plates.