In Fig. 22-42, the three particles are fixed in place and have charges q1=q2=+e and q3=+2e . Distance, a=6.0 mm. What are the (a) magnitude and (b) direction of the net electric field at point P due to the particles?
The electric field is a vector field, so the net electric field at any point can be calculated by doing the vector addition of all the fields due to the individual sources.
The electric field is given as,
where, R is the distance of field point from the charge and q is the charge on the particle
By symmetry we see that the contributions from the two charges q1=q2=+e cancel each other.
Thus, the magnitude of the net electric field is due to charge 3 which is located at a distance of from P. the magnitude of net electric field is given using equation (i) as:
Hence, the value of the net electric field is 160 N/C
As the net electric field must be directed away from the charge q3 and the length of lines 2P and 3P are equal, so the angle made by the net electric field with +x-axis is given as-
Thus, the electric field points in the direction making an angle of 45o with the positive side of the x-axis.
Two particles, each of positive charge q, are fixed in place on a y axis, one at and the other at. (a) Write an expression that gives the magnitude E of the net electric field at points on the x axis given by. (b) Graph E versus for the range. From the graph, determine the values of that give (c) the maximum value of E and (d) half the maximum value of E.
In Fig. 22-64a, a particle of charge produces an electric field of magnitude at point P, at distance R from the particle. In Fig. 22-64b, that same amount of charge is spread uniformly along a circular arc that has radius R and subtends an angle. The charge on the arc produces an electric field of magnitude at its center of curvatureP. For what value of does the electric field? (Hint: You will probably resort to a graphical solution.)
94% of StudySmarter users get better grades.Sign up for free