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Paul [167]
4 years ago
5

A positive point charge (q = +5.45 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.49 m. A p

ositive test charge (q0 = +4.96 x 10-11 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -9.05 x 10-9 J. Find rB.
Physics
1 answer:
ruslelena [56]4 years ago
5 0

Answer:

r_{B} = 3.34\ m

Solution:

As per the question:

Point charge, q = 5.45\times 10^{- 8}\ C

Test charge, q_{o} = 4.96\times 10^{- 11}\ C

Work done by the electric force, W_{AB} = - 9.05\times 10^{- 9}\ J

Now,

We know that the electric potential at a point is given by:

V = \frac{kqq'}{r}

where

r = separation distance between the charges.

Also,

The work done by the electric force i moving a test charge from point A to B in an electric field:

W_{AB} = kqq_{o}(\frac{1}{r_{B} - \frac{1}{r_{A}}}

- 9.05\times 10^{- 9} = 9\times 10^{9}\times 5.45\times 10^{- 8}\times 4.96\times 10^{- 11}(\frac{1}{r_{B}} - \frac{1}{1.49}}

- 0.3719 = (\frac{1}{r_{B}} - 0.67}

r_{B} = \frac{1}{0.299} = 3.34\ m

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Tcecarenko [31]

Answer:

The false statement is 'Electric field lines form closed loops'.

Explanation:

  • Electric field lines originate from positive end and terminates at negative end,i.e., field lines are inward in direction to the negative charges and outward from the positive charges.
  • These lines when close together represents high intensity and when far apart shows low intensity of the field.
  • These lines do not intersect, as the tangent drawn on these lines provides us with the field direction and intersection of these lines means two field directions which is not possible.
  • These lines unlike magnetic field lines do not form closed loops as they do not turn around but originate at positive end and terminates at negative end which ensures no loop formation.
8 0
4 years ago
What is the safety rules?
Black_prince [1.1K]

Answer:

Quick question do you mean what are some safety rules

Explanation:

Crosswalk, Stop sign,

5 0
4 years ago
Read 2 more answers
Finish A 16 N force is applied to an object and 96 J of work is done. How far was the object moved?
Lina20 [59]

Answer:

\boxed {\boxed {\sf 6 \ meters}}

Explanation:

Work is the product of force and distance.

W=F*d

We know that 96 Joules of work were done and a 16 Newton force was applied to the object.

  • W= 96 J
  • F= 16 N

Substitute the values into the formula.

96 \  J= 16 \ N * d

First, let's convert the units. This will make cancelling units easier later in the problem. 1 Joule (J) is equal to 1 Newton meter (N*m), so the work of 96 Joules equals 96 Newton meters.

96 \ N*m= 16 \ N * d

Now, solve for distance by isolating the variable, d. It is being multiplied by 16 Newtons and the inverse of multiplication is division. Divide both sides of the equation by 16 N.

\frac {96 \ N*m}{16 \ N}= \frac{16 \ N *d}{16 \ N}

\frac {96 \ N*m}{16 \ N}=d

The units of Newtons cancel.

\frac {96}{16} \ m = d

6 \ m = d

The object moved a distance of <u>6 meters.</u>

3 0
3 years ago
A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
tekilochka [14]

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wave
Ann [662]

Answer:

\lambda = 40 cm

Explanation:

given data

string length = 30 cm

solution

we take here equation of length that is

L = n \times \frac{1}{4} \lambda     ...............1

so

total length will be here

L = \frac{\lambda}{2} +  \frac{\lambda}{4}\\

L = \frac{3 \lambda }{4}

so \lambda  will be

\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}

\lambda = 40 cm

5 0
3 years ago
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