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svetoff [14.1K]
3 years ago
10

The following picture displays a map of potential difference (vertical axis) for an unknown configuration of charges as a functi

on of x- and y-coordinates (horizontal axes). What is the force experienced by a charged particle placed in the electric field of these charges along

Physics
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

Direction 1: Force is Non-zero and Not- constant

Direction 2: Force is Non-zero but constant  

Explanation:

Given:

The picture of the map is attached. ( Missing from the question ).

Find:

The effect of force as it travels along each direction.

Solution:

- We know the relationship between change in potential and the force acting on the charge particle is given by:

                               F = - q*dV/ dr

Where,

q : Charge of the particle

V : Volt potential

dV/dr : Potential difference along a direction.

Direction 1:

- The color code of the map changes as the particle moves along this direction. Each color code represents a potential difference. So as the particle moves between different potential difference then according to the relationship given above The force varies along varies as particle moves from one color to another. Hence, a non zero force but not constant.

Direction 2:

- In the direction 2, the charged particle moves along the same color. The potential difference for each color is constant. Hence, according to the relationship of potential difference and force. If potential difference is constant then the Electrostatic Force on the charge is also constant. Hence, Force is non-zero and constant.

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Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral
Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit  such that

P=VI=V^2/R

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

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Current 1=V/R

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Power1=P

At Bulb 1 and Bulb 2

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Current2=\frac{V}{2R}

Power2=Voltage * Current2

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2 years ago
A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
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Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

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hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

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