Physics: Atest charge is placed at a distance of 2.5 × 10-2 meters from a charge of 6.4 × 10-5 coulombs. what is the electric field at the test charge? (k = 9.0 × 109 newton·meter2/coulomb2) a. 1.6 × 10-8 newtons/ coulomb b. 2.4 × 108 newtons/ coulomb c. 3.6 × 108 newtons/coulomb d. 4.2 × 108 newtons/coulomb e. 9.2 × 108 newtons/coulomb

3.1 the only reason i know this is cause i got it wrong ...

Atest charge is placed at a distance of 2.5 ×

laycihyatt11

10.03.2022

The correct answer option is E. 9.2 × 108 newtons/coulomb.

Explanation:

We know the formula for the electric field:

E = kQ / d²

where E = electric field,

Q = the test charge in coulomb; and

d = distance.

Substituting the given values in the above formula to get:

$E=\frac{(9 * 10^9)(6.4 * 10^-5)}{(2.5 * 10^-2)^2}$

$E=\frac{576000}{6.25*10^-4}$

E=921600000 or 9.2 * 10^8 Newtons/Coulomb

Therefore, the electric field at the test charge is 9.2 * 10^8 Newtons/Coulomb.

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