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skad [1K]
3 years ago
10

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 t

he test charge? (k = 9.0 × 109 newton·meter2/coulomb2)
Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance. 

E = kQ / d²

k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m

Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²

Electric field at the test charge is 921600000 N/C
Sveta_85 [38]3 years ago
8 0

Answer:

E.  9.2 × 108 newtons/coulomb.

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<h3>Equivalent resistance of the series resistors</h3>

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<h3>Effective resistance of the circuit</h3>

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Learn more about resistors in parallel here: brainly.com/question/15121871

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