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What is the electrostatic force between a charge with +2.0 x 10 -7 C and a charge of    +1.0 x 10 -7 C separated by a distance of 0.05 meters?
in Grade 12 Physical Sciences by Platinum (110k points) | 17 views

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F=8.99X10^9 X 2X10^-7X1X10^-7/0.5^2

=(8.99X2X1) X10^9-7-7/0.5^2

=((8.99X2X1)X10^9-7-7))/2.5X10^-3

= 7.192X10^9-7-7+3

= +7.192x10^-2N
by Wooden (3.4k points)
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Fe=kq1q2/r²

= (8.99x10^9)(2.0x10^-7)(1.0x10^-7) / (0.05)²

= +0.07192 N
by Wooden (2.2k points)
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F = (Kq1q2) / r^2

   =8.99*10^9 X  2.0*10^-7 X 1.0*10^-7) / 0.05^2

    =+0.0719 N
by Wooden (1.1k points)
edited by
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Fe=kq1q2/r²

F= (8.99x10^9)(2.0x10^-7)(1.0x10^-7) / (0.05)²

F= 0.07192 N
by Wooden (426 points)
1 like 0 dislike
F = kq1q2/r^2

F = (8.99x10^9)(2.0x10^-7)(1.0x10^-7)/(0.05)^2

F = +0.07192 N
by Wooden (2k points)
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F = kq1q2/r^2

   = ( 8.99 x 10^9 )(+2.0 x 10 ^-7 )(+1.0 x 10 ^-7) / 0.05

   =  (8.99 X 2.0 X1.0) 10^9+(-14) / 0.05 ^2

   = 17.98 X 10^-5 /  0.0025

   = 0.0001798 / 0.0025

   = 0.07192 N
by Wooden (381 points)
1 like 0 dislike
Fe=Kq1q2/r^2

(8.99x10^9)x(2.0x10^-7)x(1.0x^-7)/0.05^2 =0.07192N
by Wooden (699 points)

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