Solved University-Physics Problems

Physics 3:  Introduction to Electricity & Magnetism 

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Javier Montenegro Joo


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(1) Four equal charges of magnitude Qo are placed equidistant from each other on the edge of a circumference of radius R.  Calculate the resultant electric field at a point on the perpendicular axis of the circumference and at a distance R from the plane of the circumference.


First, find only the resultant electric field from a pair of symmetrical charges, for example those two on the vertical, at the end the total electric field due to the four charges will be the double of this result. The sketch shows the situation for the vertical pair of charges, it can be seen that the vertical components of the electric fields (E1Y and E2Y) of these two charges are equal and cancel each other, hence the resultant field of this pair is the sum of their horizontal components, which are also equal (E1x = E2x):

Physics Virtual Lab (PVL):  

Electrically-charged particle deflection by an electric field: the user sets the electric field intensity, the mass of the particle, its electric charge and its initial velocity, the module executes the simulation.

The PVL has been created by Javier Montenegro Joo

Download  Physics Virtual Lab (PVL)  Demo 


(2) In the fig., a long and straight wire having a linear charge density is placed in front, parallel and very close to a sheet of paper having a surface charge density. The short distance from wire to paper is b. Determine the distance r measured from the wire, where a point charge Q does not experiment any force.


An electric charge placed between the wire and the plane will not experiment any force, if the electric field of the wire equals that of the plane:

(3) In the sketch, a charge Q is uniformly distributed along a circumference arc of radius b, which spans an angle from 45to 135o.  Determine the electric field at the origin of coordinates.


In order to solve this problem, consider a differential element of charge at the arc, sketch its electric field vector at the origin of coordinates and show its vector components along the x and y directions. Then repeat the procedure for another differential element of charge at a symmetric position in the arc. The result is that the components along the x-direction cancel each other, while the components along the y-direction add together producing the resultant electric field, which is downwards.  

(4)  Four electric charges,   5Qo ,   -4Qo ,  6Q and  Q,  are placed each in a corner of a square of side 2L.  Determine:  The charge Q, so that no electric field is detected at the center of the square.


(5)  In the figure, a wire has been bent so that it has the shape of a semi-circumference of radius R, then the wire has been loaded with an electric charge and, this distributes as a linear density  L=LoSin(Theta). Determine the electric field at the origin of coordinates.


Integrating the electric field between 0o  and 180o  (from top to bottom of the sketched arc) generates a weird result, which is neither in agreement with the symmetry of the Sine function between the mentioned points, nor with the symmetry of the arc.

The shape of the Sine function between 0o and 180, shows that the function decreases at both extremes, this is, top and bottom of the arc and, it takes a maximum at 90o , which in this sketch is around the x-axis.   Thus by an analysis of symmetry the resulting electric field must be in the x-direction.

Consider a differential charge dQ in upper region of the arc, sketch its Ex and Ey at the origin of coordinates. Then do the same for a symmetrical dQ in the lower region of the arc, it can be seen that the vertical components Ey of both dQ cancel each other and only survive the Ex components of both dQ, which must be integrated.


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