从离散电荷解净电力

Important Equations:

$F_e = \frac{Kq_1q_2}{r^2}$

$E = \frac{F}{q_o} = \frac{Kq}{r^2}$

Example 1:

Step 1: Calculate the E caused by each charge (A, B, C, D). Ignore sign while calculating the magnitude of E and use it to determine the direction of E.

$E$ from A = $\frac{(9\times10^9)(2\times10^{-9})}{(0.5)^2} = 72\frac{N}{c}(-i)$

Note that since P has a negative charge and A has a positive charge, the E from A will move in the $-i$ direction toward P. ($72\frac{N}{c}(-i)$ is the same as $-72\frac{N}{c}i$)

$E$ from B = $\frac{(9\times10^9)(4\times10^{-9})}{(1)^2} = 36\frac{N}{c}(+i)$

Note that since P has a negative charge and B has a negative charge, the E from B will move in the $+i$ direction away from P

$E$ from C = $\frac{(9\times10^9)(4\times10^{-9})}{(1.5)^2} = 16\frac{N}{c}(-i)$

$E$ from D = $\frac{(9\times10^9)(6\times10^{-9})}{(2)^2} = 13.5\frac{N}{c}(+i)$

Step 2: Add up the E at point P from A, B, C and D

$E = -72+36-16+13.5 = -38.5\frac{N}{c}i$

Step 3: Calculate the Electric Force using the E at point P from A, B, C and D

$F = q_oE = (-3\times10^{-9})(-38.5) = 1.55\times10^{-7}N$

Example 2:

$E$ from A = $\frac{(9\times10^9)(2\times10^{-9})}{(0.5)^2} = 72\frac{N}{c}i$

$E$ from B = $\frac{(9\times10^9)(2\times10^{-9})}{(0.5)^2} = 72\frac{N}{c}i$

$E$ from C = $\frac{(9\times10^9)(6\times10^{-9})}{(1)^2} = -54\frac{N}{c}i$

$E$ from D = $\frac{(9\times10^9)(2\times10^{-9})}{(0.5)^2} = -72\frac{N}{c}j$

$E$ from E = $\frac{(9\times10^9)(4\times10^{-9})}{(0.5)^2} = -144\frac{N}{c}j$

$E$ from F = $\frac{(9\times10^9)(6\times10^{-9})}{(1)^2} = 54\frac{N}{c}j$

$E = 72i + 72i - 54i - 72j - 144j +54j = 90i - 162j \frac{N}{c}$

$F = q_oE = (3\times10^{-9})( 90i - 162j) = 2.7\times10^{-7}i - 4.86\times10^{-7}j N$

In short, electric field at any point of space is the vector sum of the fields at that point of all individual sources for superposition.

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