This is just so I can flip the picture

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

Consider a W21x93. Determine the moment capacity of the beam. Assume the compression flange is not laterally braced and that the

OLga [1]

Answer:

The answer is "828.75"

Explanation:

Please find the correct question:

For W21x93 BEAM,

$Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3$

For A992 STREL,

$F_y= 50\ ks$

Check for complete section:

$\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}$

Design the strength of beam = $\phi_b Z_x F_y\\\\$

$=0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\$

8 0

3 years ago

A construction crew lifts approximately 400 lb. of material several times during a day from a flatbed truck to a 25 ft. rooftop.

Irina18 [472]

Answer:

2ib

Explanation:

if you divide 10 divided by 2 it gives you 5 and then subtract it by 2.2 = 2.8

there goes your answer.

5 0

2 years ago

1. There are two categories (shapes) for the Virginia driver's license. The ________________________ shape license represents th

dolphi86 [110]

Answer:

Vertical; horizontal.

Explanation:

The Virginia Department of Motor Vehicles started issuing sets of newly designed driver's licenses to drivers in 2009. Although, the cards that were issued to drivers prior to the introduction of the new cards remained valid until they were expired.

There are two categories (shapes) for the Virginia driver's license. The vertical shape license represents the driver who is under the age of 21, and the horizontal shaped license represents the driver who is over the age of 21.

Additionally, the Virginia's driver license (vertical in shape) issued to drivers who are under the age of 21 has a background image of a dogwood flower while the horizontal shaped license issued to drivers who are over the age of 21 has a background image of the state capitol.

5 0

3 years ago

A buret is a device designed to precisely dispense liquids. The scale is calibrated in mL with the zero point at the top with nu

kobusy [5.1K]

Answer:

As there was no attached picture, I will explain how to take the measurement of liquids in any buret which you can then apply to the specific question

Explanation:

A buret is a laboratory apparatus used to precisely measure the volume of liquids (usually alkalise or bases) used in a titration experiment. The standard buret has a capacity of 50 ml and graduated in 0.1ml though burets with smaller capacities exist.

From the question, your buret is filled to the top (0.00ml) with liquid. It is very important when taking buret readings to place the buret below your eye level so that the bottom meniscus (lower part of the liquid) can be read.

To take the buret reading, note your initial buret reading (in this case 0.00ml) then titrate the liquid base in the buret against the acid by opening the tap located at the bottom of the buret.

When the titration or reaction is complete, note the final reading against the calibration of buret. You can do this by observing the lower meniscus of the liquid remaining in the buret. (Remember to keep the buret at eye level to avoid parallax error),

The difference between your final buret reading and the initial buret reading gives you the precise volume of liquid used in the reaction.

6 0

3 years ago