Exercise one, Physics Help. Electric Field
Created by Point Charges.
Find the electric field at point P on the y axis and at a distance 0.4 m above the origin. The electric field is created by the three point charges as shown below. The charge q1 = 7x10-6 C is in the origin of the coordinate system, the charge q2 = -5x10-6 C is placed on the x axis 0.3 m from the origin and the charge q3 = -3x10-6 C is placed to the right of point P and at 0.4 m above q2.
Determine also the force exerted on a charge of 3x10-6 C when placed on the point P.
Let's first calculate the magnitude of the electric field at P due to the presence of each charge. Denote E1 the electric field produced by q1, E2 the electric field produced by q2 and E3 the electric field produced by q3. These electric fields are represented in the figure and their magnitudes are, using positive values for the charges:
E1 = keq1/r12 = (9x109 Nm2/C2)(7.0x10-6 C)/(0.4m)2
= 3.9x105 N/C.
E2 = keq2/r22 = (9x109 Nm2/C2)(5.0x10-6 C)/(0.5m)2
= 1.8x105 N/C.
E3 = keq3/r32 = (9x109 Nm2/C2)(3.0x10-6 C)/(0.3m)2
= 3.0x105 N/C.
Vector E1 has no x component, just y component (upward). Vector E2 has a x component given by E2cos▀ = (3/5)E2 and a negative y component given by -E2 sen▀ = -(4/5)E2. Vector E3 has no y component, just x component (to the right).
resultant vector E we are looking for is the vectorial sum of this three vectors,
E = E1 + E2 + E3. The vectors E1,
E2 and E3 are denoted using unitary vectors i and j.
Then we can obtain their sum analytically.
E1 = 3.9x105 j N/C.
E2 = (1.1x105 i - 1.4x105 j )N/C.
E3 = 3.0x105 i N/C.
The electric field E at P is then:
E = (4.1x105 i + 2.5x105 j) N/C.
The force on a charge of 3x10-6 C when is placed at the point P is obtained using F = Eq, with q = 3x10-6 C, F = (1.23 i + 0.75 j)N. This force has of course the same direction of the electric field E, because the charge q is positive.
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Physics Help, Exercise Two, Electric Field.
In the figure a thin rod of length L and a total charge Q uniformly distributed is placed along the x axis. Find the magnitude of the electric field at point A located a distance d from the left side of the rod.
This problem can be solved assuming the rod is made of point charges of magnitude dq = (Q/L)dx. Q/L is the charge per unit length. We can use E = Kq/x2 in differential form, dE = Kdq/x2. dE is the contribution to the field of dq at a distance x from A. Then we add the contributions of all the electric point charges using limits of integration from d to d + L:
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