A very long l, straight wire has charge per unit length, 1.5×10^-10 C/m. At what distance from the wire is the is the electric field magnitude equal to 2.50N/C? ?

To find the distance from the wire where the electric field magnitude is equal to 2.50 N/C, we can use the formula for the electric field created by a long straight wire:

E = (k * λ) / r

where:
E = electric field magnitude (2.50 N/C)
k = Coulomb's constant (8.99 x 10^9 N m^2/C^2)
λ = charge per unit length (1.5 x 10^-10 C/m)
r = distance from the wire

Plugging in the values, we get:

2.50 = (8.99 x 10^9 * 1.5 x 10^-10) / r

Solving for r:

r = (8.99 x 10^9 * 1.5 x 10^-10) / 2.50
r = 5.394 x 10^-1 m
r = 0.5394 m

Therefore, the distance from the wire where the electric field magnitude is equal to 2.50 N/C is 0.5394 meters.

The electric field magnitude of a very long, straight wire with charge per unit length can be calculated using the formula:

$$E = \frac{2kq}{r}$$

Where:

- E is the electric field magnitude
- k is Coulomb's constant ($8.9876 \times 10^9 N \cdot m^2/C^2$)
- q is the charge per unit length
- r is the distance from the wire

We are given that the charge per unit length is $1.5 \times 10^{-10} C/m$ and the electric field magnitude is 2.50 N/C. We need to find the distance from the wire.

Rearranging the formula to solve for r, we get:

$$r = \frac{2kq}{E}$$

Plugging in the given values, we get:

$$r = \frac{(2)(8.9876 \times 10^9 N \cdot m^2/C^2)(1.5 \times 10^{-10} C/m)}{2.50 N/C}$$

$$r = 0.2696 m$$

Therefore, the distance from the wire where the electric field magnitude is 2.50 N/C is 0.2696 m.