1. How large is the electric field (magnitude only) at a point that is 40.0 cm a

Question

1. How large is the electric field (magnitude only) at a point that is 40.0 cm away from an isolated 1.94×105 C charge?

[Answer: 1.09 e6 N/C is WRONG because it wants the answer in dimensions of electric field!]

2. A downward electric force of 7.2N is exerted on a 6.4C charge.Find the magnitude of the electric field at the position of this charge.

[Answer: 1.125e6 N/c is WRONG because it wants the answer in dimensions of electric field!]

****SHOW WORK IN A CLEAR FORMAT PLEASE! I WANT TO UNDERSTAND HOW YOU GOT THE SOLUTION! THANKS! also, ***ENTER ANSWER USING DIMENSIONS OF ELECTRIC FIELD***

Explanation / Answer

1. electric field from point charge = (charge) / [(4o)(distance from charge)^2]

E = (1.94x10^-5 C) / [(4*8.854x10^-12 C^2/N-m^2)(0.40 m)^2]
E = 1.089x10^6 N/C

the unit of electric field is N/C..

2.

F = q*E

E = F/q = 7.2 N downward / -6.4 uC = -1.125 * 10^6 N/C downward

Since it is "negative" downward, that means "the opposite of" downward, or upwards.

So the answer really is
E = 1.125x10^6 N/C upward

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