Answer:1.008 ×10^-14/rJ
Where r is the distance from.which the charge was moved through.
Explanation:
From coloumbs law
Work done =KQq/r
Where K=9×10^9
Q=7×10^-6C
q=e=1.6×10^-19C
Micro is 10^-6
W=9×10^9×7×10^-6×1.6×10^-19/r=100.8×10^-16/r=1.008×10^-14/rJ
r represent the distance through which the force was used to moved the charge through.
Complete question:
A 7.0 -μC point charge and a 9.0 -μC point charge are initially infinitely far apart. How much work does it take to bring the 7.0-μC point charge to x=3.0 mm, y= 0.0mm and the 9.0-μC point charge to x=-3.0 mm, y=0.0 mm? (the value of k is 9.0*10^9 N*m^2/C^2)
Answer:
Explanation:
Work done in bringing a unit positive charge from infinity to that point in electric filed (V) = Electric field strength X distance
[tex]V = EXd = \frac{kq}{d^2}Xd=\frac{kq}{d}[/tex]
where;
K is a constant = 9X10⁹ N.m²/C²
q is point charge in C
d is the distance in m
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Work done in bringing 7.0-μC point charge from infinity to x=3.0 mm, y= 0.0mm
[tex]V = \frac{(9X10^9) X(7X10^{-6})}{3x10^{-3}}[/tex]
V = 21 x 10⁶ Volts
Work done in bringing 9.0-μC point charge from infinity to x=-3.0 mm, y=0.0 mm
[tex]V = \frac{(9X10^9) X(9X10^{-6})}{-3x10^{-3}}[/tex]
V = -27 x 10⁶ Volts
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Work done in bringing 7.0-μC point charge from infinity to x=3.0 mm, y= 0.0mm is 21 x 10⁶ Volts While work done in bringing 9.0-μC point charge from infinity to x=-3.0 mm, y=0.0 mm is -27 x 10⁶ Volts.
