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3 years ago

Which of the following represent units of an electric field? select all that apply

Physics

2 answers:

zimovet [89]3 years ago

8 0

Answer:

Volts/Meter

Newtons/Coulomb

Explanation:

Volts/ Meter and Newtons/Coulomb both are same and the units of Electric field intensity or electric field strength.

Electric field strength E is the force per unit charge. It is measured in Newton/Coulomb in SI unit. It is a vector quantity directed in the direction of force.

Mathematically,

Electric field strength = Force/Charge

E = F / q₀

= Newton / Coulomb = NC⁻¹ 1

We know that

Newton = Joule/meter 2

Also

Volt = Joule/Coulomb 3

So put 3 in 2 we get

Newton = (Volt Coulomb)/meter put in 1

E = (Volt Coulomb)/(meter Coulomb)

= Volt / meter

Hence

Newton / Coulomb = Volt / meter

blsea [12.9K]3 years ago

4 0

Answer:

Volts/Meter

Newtons/Coulomb is correct i got it right on odyssey

Explanation:

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An electric vehicle starts from rest and accelerates at a rate a1 in a straight line until it reaches a speed of v. The vehicle

levacccp [35]

(a) $t=\frac{v}{a_1}+\frac{v}{a_2}$

In the first part of the motion, the car accelerates at rate $a_1$ , so the final velocity after a time t is:

$v = u +a_1t$

Since it starts from rest,

u = 0

So the previous equation is

$v= a_1 t$

So the time taken for this part of the motion is

$t_1=\frac{v}{a_1}$ (1)

In the second part of the motion, the car decelerates at rate $a_2$ , until it reaches a final velocity of v2 = 0. The equation for the velocity is now

$v_2 = v - a_2 t$

where v is the final velocity of the first part of the motion.

Re-arranging the equation,

$t_2=\frac{v}{a_2}$ (2)

So the total time taken for the trip is

$t=\frac{v}{a_1}+\frac{v}{a_2}$

(b) $d=\frac{v^2}{2a_1}+\frac{v^2}{2a_2}$

In the first part of the motion, the distance travelled by the car is

$d_1 = u t_1 + \frac{1}{2}a_1 t_1^2$

Substituting u = 0 and $t_1=\frac{v}{a_1}$ (1), we find

$d_1 = \frac{1}{2}a_1 \frac{v^2}{a_1^2} = \frac{v^2}{2a_1}$

In the second part of the motion, the distance travelled is

$d_2 = v t_2 - \frac{1}{2}a_2 t_2^2$

Substituting $t_2=\frac{v}{a_2}$ (2), we find

$d_1 = \frac{v^2}{a_2} - \frac{1}{2} \frac{v^2}{a_2} = \frac{v^2}{2a_2}$

So the total distance travelled is

$d= d_1 +d_2 = \frac{v^2}{2a_1}+\frac{v^2}{2a_2}$

7 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm

mihalych1998 [28]

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?

Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.

(a) What is the electric potential at point a due to q1 and q2?

(b) What is the electric potential at point b?

(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Answer:

a) the potential is zero at the center .

Explanation:

a) since the two equal-magnitude and oppositely charged particles are equidistant

b)(b) Electric potential at point b, v = Σ kQ/r

r = 5cm = 0.05m

k = 8.99*10^9 N·m²/C²

Q = -2 microcoulomb

v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m

v = -105 324 V

c)workdone = charge * potential

work = -6.00µC * -105324V

work = 0.632 J

6 0

3 years ago

What equation describes conservation of charge?

Phantasy [73]

Answer:

The equation which describes conservation of charge is $Q_{initial} - Q_{final } = 0$

Explanation:

The law of conservation charge states that for an isolated system that sum of initial charges is equal to sum of final charges, that is the total charge is conserved.

let the sum of initial charges = $Q_{initial}$