Point charges are depicted in a figure. (I can\'t attach the figure but it is a

Question

Point charges are depicted in a figure. (I can't attach the figure but it is a square with charges located in each corner... clockwise from top left: A, B, D, C, with q in the middle). A) Calculate the magnitude of the electric field at the location of q in N/C, given that the square is 6.5 cm on each side. qa=qb= -1.4 uC and qc=qd= 1.4 uC. Answer is E in absolute value bars |E|.

Now consider the same figure when q= 0.75 uC, qa= 1.1 uC, qb= -2.6 uC, 9c= -5.8 uC, and qd= 1.1 uC. If the square is 36 cm on each side, what is the total Coulomb force (in N)? Again, F is in absolute value bars |F|.

Please explain; I am asking this question to understand the process more than the answer, thank you!

Explanation / Answer

Hi,

To find the total electric field, you must first find the electric field generated by each charge, but you have to remember that the electric field is a vectorial quantity, therefore you must add the components of such fields.

The following equation gives you the value of the electric field (its magnitude):

E = k q/r2 ; where k is the electric constant, q is the charge and r is the distance from the charge to the point of interest.

Each charge creates a field of the same magnitude, since all of them are at the same distance from the center of the square:

r = (6.5/2) cm (2)1/2 = 4.6 cm

Eo = (9*109 Nm2/C2 ) (1.4*10-6 C) / (4.6*10-2 m)2 = 5.96*106 N/m

As you have noticed, the signs of the charges were not considered. That is because they matter once we consider the direction of the field they create.

- Charges at A and B would attract a positive trial charge, so the field they create is directed towards them.

- Charges at C and D would repel a positive trial charge, so the field they create is directed in the opposite direction of them.

However, the charges at A and D would move the charge to the left, while the charges at B and C would move the charge to the right. As the field the charges create is equal in magnitude, the trial chage would not move in the x direction (so the resultant electric field does not have an horizontal component).

On the other hand, all the charges would move the trial charge upwards (in the positive y direction), so it is clear that the resultant electric field is over the y axis and it is the result of the contribution of all the charges.

So, in the end, the horizontal components of the electric fields are not requiered, since they nullified each other. On the other hand we have to calculate the vertical components as the resultant electric field is four times said value.

Eoy = sin(45°) Eoy = 4.21*106 N/m (the angle of 45° is due to geometry of the situation)

E = 4Eoy = 4(4.21*106 N/m) j = 1.68*107 N/m j

The magnitude of E is 1.68*107 N/m.

I hope it helps.

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