The electric field strength 6.0 cm from a very long charged wire is 1700 N/C. What is the electric field strength 10.0 cm from the wire?

1020 N/C

Correct. E for an infinitely long charged wire varies inversely with distance, not distance-squared.

To determine the electric field strength at a distance of 10.0 cm from the wire, we can use the principle of symmetry.

The electric field strength due to a long charged wire is inversely proportional to the distance from the wire. Mathematically, we can express this relationship as:

E ∝ 1/r

Where E is the electric field strength and r is the distance from the wire.

We are given that at a distance of 6.0 cm from the wire, the electric field strength is 1700 N/C. Let's denote this electric field strength as E1 and the distance as r1.

E1 = 1700 N/C
r1 = 6.0 cm = 0.06 m

Using the relationship E ∝ 1/r, we can write:

E1 / E2 = r2 / r1

Now, we need to rearrange the equation to solve for E2:

E2 = (E1 * r2) / r1

Plugging in the given values:

E2 = (1700 N/C * 10.0 cm) / 6.0 cm

Note that we convert the distance from centimeters to meters, as the electric field strength is commonly expressed in N/C, which is a SI unit.

E2 = (1700 N/C * 0.10 m) / 0.06 m

Calculating this expression:

E2 = (170 N*m/C) / 0.06 m

E2 ≈ 2833 N/C

Therefore, the electric field strength at a distance of 10.0 cm from the wire is approximately 2833 N/C.