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xeze [42]
3 years ago
6

A thin, metallic spherical shell of radius 0.187 m has a total charge of 6.53×10−6 C placed on it. A point charge of 5.15×10−6 C

is placed at the center of the shell. What is the electric field magnitude ???? a distance 0.965 m from the center of the spherical shell? ????= N/C Does the electric field point outward from the center of the sphere or inward?
Physics
1 answer:
MAVERICK [17]3 years ago
5 0

Answer: a) 112.88 * 10^3 N/C; b) The electric field point outward from the center of the sphere.

Explanation: In order to solve this problem we have to use the gaussian law so we use a gaussian surface at r=0.965 m and the electric flux is equal to Q inside/εo

E* 4*π*r^2= Q inside/εo

E= k*Q inside/r^2= 9*10^9*(6.53+5.15)μC/(0.965)^2=122.88 * 10 ^3 N/C

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3 years ago
Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the pos
deff fn [24]

Answer:

E = (0, 0.299) N

Explanation:

Given,

  • Charge q_1\ =\ 8.0\ nC
  • Charge q_2\ =\ 6.0\ nC
  • Distance of the first charge from the origin = (16m, 0)
  • Distance of the second charge from the origin = (-9, 0)
  • Point where the electric field required = (0, 12m)

Let \theta_1\ and\ theta_2 be the angle of the electric fields by first and second charge at the point A.

\therefore sin\theta_1\ =\ \dfrac{12}{20}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{20}\ \right )\\\Rightarrow \theta_1\ =\ 36.87^o\\\\\therefore sin\theta_1\ =\ \dfrac{12}{9}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{9}\ \right )\\\Rightarrow \theta_1\ =\ 53.13^o\\

Electric field by charge q_1 at point A,

F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 8\times 10^{-9}}{20^2}\\\Rightarrow F_1\ =\ 0.18\ N/C

Electric field by the charge q_2 at point A,

F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 6.0\times 10^{-9}}{16^2}\\\Rightarrow F_1\ =\ 0.24\ N/C

Now,

Net electric field in horizontal direction at point AF_x\ =\ F_{1x}\ +\ F_{2x}\\\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\\\Rightarrow F_x\ =\ 0.18\times( -cos36.87^o)\ +\ 0.24\times cos53.13^o\\\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\\\Rightarrow F_x\ =\ 0\ N/C

Net electric field in vertical direction at point A.

F_y\ =\ F_{1y}\ +\ F_{2y}\\\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\\\Rightarrow F_y\ =\ 0.18\times sin36.87^o\ +\ 0.24\times sin53.13^o\\\Rightarrow F_y\ =\ 0.180\ +\ 0.192\\\Rightarrow F_y\ =\ 0.299\ N/C

Hence, the net electric field  at point A,

F\ =\ ( 0, 0.299 )\ N/C.

5 0
3 years ago
A string of 18 identical holiday tree lights is connected inseries to a 190V source. The string dissipates 62.0 W. One of thebul
PtichkaEL [24]

Answer:

549.9 ohms, 65.65 watts, the power went up

Explanation:  power P = V²/R

v= 190volts , P =62watts

from P =V²/R

62=190²/R

making R the subject

R×62= 190²

62R =36100

R=36100/62

R=586.25 ohms

Resistance of each lamp = 586.26/18 =32.3ohms

a) resistance of the light string now = 17×32.3 = 549.9 ohms

b) P=V²/R

where R=549.9ohms , P=?

P=190²/549.9

P=36100/549.9

P=65.65watts

Power dissipated has increased( went up )

7 0
3 years ago
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