Consider a platinum wire (σ= 1.0 × 107 Ω-1·m-1) with a cross-sectional area of 1 mm2 (similar to your connecting wires) and carrying 0.3 amperes of current, which is about what you get in a circuit with a round bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

0.03 N/C

Explanation:

We use the current density formula to solve this question.

I/A = σ * E

Where,

I = current flowing in the circuit = 0.3 A

A = cross sectional area of the wire = 1 mm²

σ = resistivity of the wire = 1*10^7 Ω^-1·m^-1

E = strength of the electric field required

I/A = σ *E

E = I/(A * σ)

First we convert area from mm to m, so that, 1*10^-3 mm = 1*10^-6 m

E = 0.3 A / (1*10^-6 m * 1*10^7 Ω^-1·m^-1)

E = 0.3 A / 10 Ω^-1

E = 0.03 N/C

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

it's a b and