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A test 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

Respuesta :

alinakincsem

Answer:

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:

[tex]E=\frac{(9 * 10^9)(6.4 * 10^-5)}{(2.5 * 10^-2)^2}[/tex]

[tex]E=\frac{576000}{6.25*10^-4}[/tex]

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

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

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