Two point charges, q 1 = -12.5 ?C and and q 2 = +18.7 ?C, are placed at two of t

Question

Two point charges, q1 = -12.5 ?C and and q2 = +18.7 ?C, are placed at two of the corners of a right triangle with a = 23.5 cm and b = 13.0 cm, as shown in the figure.

a) Find the magnitude of the electric field at the third corner of the triangle.

b) Determine the direction of electric field at the third corner of the triangle (angle with respect to the positive x-axis).

Explanation / Answer

Hello Crockodile Dundee, let us imagine that a is along positive X axis and b along Y axis. Then as we place the charge negative at the top and positive at the right extreme of the side a. then the field at the origin due to negative will be along positive Y axis and the field due to positive charge will be along negative X direction. Hence the resultant will be inclined with acute angle with the negative X axis or with obtuse angle with the positive X axis. Ok. Now let us use the formula for field as E = k q/ r^2 For negative charge, q = 13.7 *10-6 C and r = 13.5*10^-2 m and k = 9*10^9 N m^2 C^-2 For positive charge, q = 15.1*10^-6 C and r = 24*10^-2 m. Hence the field due to negative E_ = 6.77*10^6 N/C Field due to positive E+ = 2.34*10^6 N/C Now these two field vectors are perpendicular to each other the resultant will be along the hypotenuse. HEnce resultant field Er = [./(6.77)^2 + (2.34)^2 ]*10^6 N/C =====> Er = 7.16*10^6 N/C. To know about the angle @ = arc tan (6.77/2.34) = 71 deg nearly. SO the angle with positive X axis will be 180-71 = 109 deg.

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