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3 years ago

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A charge of 6.7 × 10^-15 coulombs is located at a point where its potential energy is 5.6 × 10^-12 joules. What is the electric

potential at that point?
A. 2.3 × 10^2 volts
B. 4.7 × 10^2 volts
C. 8.4 × 10^2 volts
D. 9.2 × 10^2 volts

2 answers:

larisa86 [58]3 years ago

6 0

<h2>Answer</h2>

Option C - 8.4 × 10^2 volts

<u>Explanation</u>

Since, formula for electric potential is given by;

V= PE/ q

Here, V= electric potential

PE= potential energy

& q= charge

Given,

q= 6.7 × 10^-15 coulombs & PE= 5.6 × 10^-12 joules.

By putting the values of ‘q’ and ‘PE’ in the formula, we get:

V= 5.6 × 10^-12/ 6.7 × 10^-15

V= 8.35x10^2

V= 8.4x10^2 volts.

Therefore, the required electric potential of the charge is 8.4x10^2 volts.

natita [175]3 years ago

4 0

Answer:

C. 8.4 × 10^2 volts

Explanation:

The potential energy of a charge is given by:

$U=qV$

where

q is the magnitude of the charge

V is the electric potential

In this problem, we have

$q=6.7\cdot 10^{-15} C$ is the charge

$U=5.6\cdot 10^{-12} J$ is the potential energy

Re-arranging the formula and using these numbers, we can find the electric potential:

$V=\frac{U}{q}=\frac{5.6\cdot 10^{-12} J}{6.7\cdot 10^{-15} C}=835.8 V = 8.4\cdot 10^2 V$

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(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

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<h2>(a) Mass of gold </h2><h2 />

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

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In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

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<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

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If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

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$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

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