All categories

4 years ago

How is the electrostatic force affected when the magnitude of a charge is doubled?

1 answer:

4 0

The magnitude of the electrostatic force between two charges is given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>

You might be interested in

Aligned magnetic domains are only present in which of the following?

butalik [34]

Answer:

i think c

Explanation:

8 0

3 years ago

What is the effect of changing the wire in a circuit from a straight thick wire to a straight thin wire?

Murrr4er [49]

Its letter B, im pretty sure

3 0

3 years ago

What is the relationship among amplitude,crest,and trough?

Musya8 [376]

These are characteristics of a wave. The amplitude is how high and low the waves go. Crests are high points on the wave, and troughs are low points on the wave.

5 0

4 years ago

Which vector correctly indicates the direction of centripetal acceleration experienced by the car?

gogolik [260]

The answer is C.

I hope this helped!

5 0

3 years ago

Read 2 more answers

A gas occupies a volume of 1.0 m3 in a cylinder at a pressure of 120kPa. A piston compresses the gas until the volume is 0.25m3,

Hoochie [10]

Answer:

Approximately $480\; \rm kPa$ , assuming that this gas is an ideal gas.

Explanation:

Let $V(\text{Initial})$ and $P(\text{Initial})$ denote the volume and pressure of this gas before the compression.
Let $V(\text{Final})$ and $P(\text{Final})$ denote the volume and pressure of this gas after the compression.

By Boyle's Law, the pressure of a sealed ideal gas at constant temperature will be inversely proportional to its volume. Assume that this gas is ideal. By this ideal gas law:

$\displaystyle \frac{P(\text{Final})}{P(\text{Initial})} = \frac{V(\text{Initial})}{V(\text{Final})}$ .

Note that in Boyle's Law, $P$ is inversely proportional to $V$ . Therefore, on the two sides of this equation, "final" and "initial" are on different sides of the fraction bar.

For this particular question:

$V(\text{initial}) = 1.0\; \rm m^3$ .
$P(\text{Initial}) = 120\; \rm kPa$ .

$V(\text{final}) = 0.25\; \rm m^3$ .
The pressure after compression, $P(\text{Final})$ , needs to be found.

Rearrange the equation to obtain:

$\displaystyle P(\text{Final}) = \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial})$ .

Before doing any calculation, think whether the pressure of this gas will go up or down. Since the gas is compressed, collisions between its particles and the container will become more frequent. Hence, the pressure of this gas should increase.

$\begin{aligned}P(\text{Final}) &= \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial})\\ &= \frac{1.0\; \rm m^{3}}{0.25\; \rm m^{3}} \times 120\; \rm kPa = 480\; \rm kPa\end{aligned}$ .

4 0

4 years ago