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2 years ago

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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)?

2 answers:

vazorg [7]2 years ago

4 0

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.

PtichkaEL [24]2 years ago

4 0

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

--------------------------------------------------------------------------------------------

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

--------------------------------------------------------------------------------------------

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.

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2 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

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1 year ago

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as , where a

Misha Larkins [42]

Answer:

3 times louder

Explanation:

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Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹

and I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₁ = 10㏒(I₁/I₀)

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L₁ = 10㏒(10³)

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So, substituting the variables into the equation, we have

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Nimfa-mama [501]

Answer:

♕ $\large{ \red{ \tt{Step - By - Step \: Explanation}}}$

☃ $\underline{ \underline{ \blue{ \large{ \tt{G \: I \: V \: E\: N}}}}} :$

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~Plug the known values and then multiply!

↦ $\large{ \tt{7 \times 42}}$

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☥ $\large{ \boxed{ \boxed{ \large{ \tt{Our \: Final \: Answer : \underline{ \large{ \tt{294 \: m {s}^{ - 1}}}}}}}}}$

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2 years ago