Question

A 7.0-μC point charge and a point charge are initially extremely far apart. How much work does it take to bring the point charge to the point , and the point charge to the point ( k = 1/4 πε 0 = 9.0 × 10 9 N ∙ m 2/C 2)?

Answer 1

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.

Answer 2

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:

• Work done in bringing 7.0-μC point charge from infinity to x=3.0 mm, y= 0.0mm is 21 x 10⁶ Volts

• 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.

Explanation:

Work done in bringing a unit positive charge from infinity to that point in electric filed (V) = Electric field strength X distance

$V = EXd = \frac{kq}{d^2}Xd=\frac{kq}{d}$

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

$V = \frac{(9X10^9) X(7X10^{-6})}{3x10^{-3}}$

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

$V = \frac{(9X10^9) X(9X10^{-6})}{-3x10^{-3}}$

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.