Ask question

Ask question

All categories

2 years ago

11

A small charge q is placed near a large spherical charge Q. The force experienced by both charges is F. The electric eld created

by Q at the position of q is:

1 answer:

anastassius [24]2 years ago

7 0

The electric field created by Q at the position of q is $\frac{F}{Q}$ .

The given parameters:

<em>Magnitude of charge, = q</em>
<em>Spherical charge, = Q</em>
<em>Force experienced by both charges, = F</em>

The electric field created by Q at the position of q is calculated as follows;

$E = \frac{F}{Q} \\\\$

where;

<em>E is the magnitude of electric field strength </em>
<em>F is the force experienced by both charges</em>
<em>Q is the charge</em>

Thus, the electric field created by Q at the position of q is $\frac{F}{Q}$ .

Learn more about electric field here: brainly.com/question/14372078

You might be interested in

The upward force exerted on an object falling through air is _____.

stira [4]

(C) Air Resistance

<u>Explanation:</u>

When an object falls through air, air resistance acts on it in upward direction. When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its surface area. Increasing the surface area of an object decreases its speed.

6 0

3 years ago

VMariaS [17]

Answer:

a.proton, proton

hope this helped :) have a goodday

6 0

3 years ago

A group of runners complete a 26.2 mile marathon in 3.4 hours. The distance between the start and finish lines is 12.2 miles. Wh

NNADVOKAT [17]

26.2/3.4 would be the average velocity for the run.

7.7 miles/hr

8 0

2 years ago

Read 2 more answers

A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T

sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

$\Delta U = Q-W$

$\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)$

Where

$\dot{m} =$ Mass flow

$h_1 =$ Specific Enthalpy (IN)

$h_2 =$ Specific Enthalpy (OUT)

$g =$ Gravity

$z_{1,2} =$ Heigth state (In, OUT)

$V_{1,2} =$ Velocity (In, Out)

Our values are given by,

$T_i = 10\°C$

$P_1 = 100kPa$

$\dot{m} = 5kg/s$

$z_2 = 20m$

For this problem we know that as pressure, temperature as velocity remains constant, then

$h_1 = h_2$

$V_1 = V_2$

Then we have that our equation now is,

$\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}$

$\dot{W} = \frac{(5)(9.81)(0-20)}{1000}$

$\dot{W} = -0.98kW$

8 0

3 years ago

How much time will it take to do 33j of work with 11 w of power

Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

$\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s$

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

$\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}$

8 0

3 years ago