
#1
Jan1108, 09:37 PM

P: 452

1. The problem statement, all variables and given/known data
A small object A, electrically charged, creates an electric field. At a point P located 0.250m directly north of A, the field has a value of 40.0N/C directed to the south. A) 1.11×10^−9C B) −1.11×10^−9C C) 2.78×10^−10C D) −2.78×10^−10C E) 5.75×10^12C F) −5.75×10^12C What is the charge of object A? 2. Relevant equations E = (KQ)/r^2 3. The attempt at a solution After manipulating the above equation, I found out that the charge Q = (E*r^2)/K Q = (E*r^2)/K = (40*(0.25^2))/9×10^9 = 2.78×10^−10C makes sense to me, but should it be negative or positive? 



#2
Jan1108, 10:16 PM

HW Helper
P: 1,664





#3
Jan1108, 10:20 PM

P: 662

A way to remember the rule that dynamicsolo stated is to recall that E fields are defined in terms of the force felt by a unit test charge, which is positive by assumption. That gives the directions stated above.




#4
Jan1108, 10:24 PM

HW Helper
P: 1,664

Magnitude and Direction of Electric Fields 



#5
Jan1108, 11:57 PM

P: 452

so it would be positive?




#6
Jan1208, 12:21 AM

Sci Advisor
HW Helper
Thanks
P: 25,161





#7
Jan1208, 07:53 AM

P: 662

cse63146  Think it through in these steps:
 Is the test charge (i.e. field point in question) North or South of the field source at A?  Given that location, does the direction of the force it feels indicate an attractive or repulsive force?  Considering that the test charge is defined to be positive, what does your previous answer tell you about the sign of the source charge at A? 



#8
Jan1208, 09:28 AM

P: 452

Since there's a point P north of charge A, and the electric field is directed to the south (towards the charge), it looks like it's inward, which makes the charge negative?




#9
Jan1208, 11:58 AM

HW Helper
P: 1,664





#10
Jan1208, 12:25 PM

P: 452

There's a part B of the question which only showed up after you answer part A.
If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P? I know that E = (KQ)/r^2 so E = (KQ)/r^2 E = ((9×10^9)(2.78×10^−10)/0.5^2 E = 10 So the total magnitude produced by the two objects 10 + 40 = 50N/C? 



#11
Jan1208, 12:34 PM

P: 3





#12
Jan1208, 12:42 PM

P: 452

Got both questions right. Thank you all.



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