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What magnitude point charge creates a 9,000 N/C electric field at a distance of 1.00 m? a. 1.00 C b. 1.00 mC c. 1.00 ?C d. 1.00 nC

Sagot :

Eduard22slyid

Answer:

1 µC

Explanation:

From the question given above, the following data were obtained:

Electric field intensity (E) = 9000 N/C

Distance (r) = 1 m

Charge (Q) =?

NOTE: Electric force constant (K) = 9×10⁹ Nm²C¯²

The magnitude of the point charge can be obtained as follow:

E = KQ/r²

9000 = 9×10⁹ × Q / 1²

9000 = 9×10⁹ × Q

Divide both side by 9×10⁹

Q = 9000 / 9×10⁹

Q = 1×10¯⁶ C

Recall:

1 micro charge (µC) = 1×10¯⁶ C

Hence, the magnitude of the point charge is 1 µC

Abidemiokinid

Answer:

1ηC

Explanation:

The magnitude of the point charge is expressed as;

V = kQ/r

k is the coulombs constant = 9.0×10^9 N⋅m²/C²

V is the electric field = 9000N/C

Q is the charge

r is the distance = 1.00m

Get the required charge

Substitute the given parameters into the formula;

9000 = 9.0×10^9Q/1.0

9000 = 9.0×10^9Q

Q = 9000/9.0×10^9

Q = 9.0×10^3/9.0×10^9

Q = 9.0/9.0×10^(3-9)

Q = 1.0×10^6C

Q = 1ηC

hence the magnitude of the point charge is 1ηC

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