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3 years ago

Does the zero electric field intensity in a given region imply zero potential?

Physics

1 answer:

Rama09 [41]3 years ago

7 0

Answer:

No, just because the electric field is zero at a particular point, it does not necessarily mean that the electric potential is zero at that point. ... At the midpoint between the charges, the electric field due to the charges is zero, but the electric potential due to the charges at that same point is non-zero.

Explanation:

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What is terminal velocity and how is it reached?

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3 years ago

In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire

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2 years ago

Read 2 more answers

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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3 years ago

What happens to the gravitational force between two objects when their distance is changed?

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If they become closer, it is increased, and if the objects become farther away is decreased.

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2 years ago

With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by

lara31 [8.8K]

According to the net force, the acceleration of the book is 16.47 m/s².

We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as

∑F = m . a

where F is force, m is mass and a is acceleration

From the question above, we know that

m = 3 kg

g = 9.8 m/s²

F1 = 20 N

Find the net force

∑F = F1 + W

∑F = 20 + m . g

∑F = 20 + 3 . 9.8

∑F = 20 + 29.4

∑F = 49.4 N

Find the acceleration

∑F = m . a

49.4 = 3 . a

a = 16.47 m/s²

Find more on force at: brainly.com/question/25239010

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7 months ago

Does the zero electric field intensity in a given region imply zero potential?​

Does the zero electric field intensity in a given region imply zero potential?