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love history [14]
3 years ago
7

Two small insulating spheres with radius 3.50×10^−2m are separated by a large center-to-center distance of

Physics
1 answer:
Liula [17]3 years ago
7 0

Answer:

Part (i) the magnitude E of the electric field midway between the spheres is 8.71 × 10⁵ N/C

Part (ii) the direction of the electric field is towards the negative charge

Explanation:

Given;

+q = 3.93.90μC, r = 3.50×10⁻²m

-q = −2.40μC, r = 3.50×10⁻²m

magnitude of the electric field is experienced, midway between the spheres at a distance r,  r = ¹/₂ × 0.51 = 0.255 m

Electric field due to point charge is given as;

E = \frac{F}{q} = \frac{Kq^2}{qr^2}  = \frac{kq}{r^2}

K is coulomb's constant = 8.99 x 10⁹ Nm²/C²

The positive charge on positive x-axis and the negative charge is on negative x-axis.

part (a)

The electric field due to positive charge; +q = 3.93.90μC

E_+ = \frac{kq}{r^2}\\\\E_+ = \frac{8.99 X10^9*3.9X0^{-6}}{0.255^2}\\\\E_+= 5.3919 X10^5\frac{N}{C}

The electric field due to negative charge; -q = −2.40μC

E_- = \frac{kq}{r^2}\\\\E_- = \frac{8.99 X10^9*2.4X0^{-6}}{0.255^2}\\\\E_-= 3.3181 X10^5\frac{N}{C}

From superimposition theorem

The magnitude of the electric field is;

E = E₊ + E₋

E = (5.3919 × 10⁵ + 3.3181 × 10⁵) N/C

E = 8.71 × 10⁵ N/C

Therefore, the magnitude E of the electric field midway between the spheres is 8.71 × 10⁵ N/C

Part (b)

The direction of the electric field is towards the negative charge.

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
3 years ago
A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long
noname [10]

Answer:

Part a)

\phi = 2.76 \times 10^{-7} T m^2

Part B)

M = 5.52 \times 10^{-5} H

Part C)

EMF = 0.1 V/s

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

B = \mu_0 n i

n = number of turns per unit length

n = \frac{N}{L}

n = \frac{350}{0.20}

n = 1750 turn/m

now we know that magnetic field due to solenoid is

B = (4\pi \times 10^{-7})(1750)(0.100)

B = 2.2 \times 10^{-4} T

Now magnetic flux due to this magnetic field is given by

\phi = B.A

\phi = (2.2 \times 10^{-4})(\pi r^2)

\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)

\phi = 2.76 \times 10^{-7} T m^2

Part B)

Now for mutual inductance we know that

\phi_{total} = M i

\phi_{total} = N\phi

\phi_{total} = 20(2.76 \times 10^{-4})

\phi_{total} = 5.52 \times 10^{-6}

now we have

M = \frac{5.52 \times 10^{-6}}{0.100}

M = 5.52 \times 10^{-5} H

Part C)

As we know that induced EMF is given as

EMF = M \frac{di}{dt}

EMF = 5.52 \times 10^{-5} (1800)

EMF = 0.1 V/s

3 0
3 years ago
PLEASE HELP!!!
Simora [160]
I only know what number 1. is and its Mechanical Energy.
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3 years ago
A jet ski accelerates towards a ramp at 2.5 m/s/s for 35 s until it finally flies off the water. Determine the
zlopas [31]

Answer:

1531 m

Explanation:

The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For the jet ski in this problem,

a=2.5 m/s^2

t = 35 s

u = 0 (it starts from rest)

Solving for s, we find the distance travelled:

d=0+\frac{1}{2}(2.5)(35)^2=1531 m

8 0
3 years ago
A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of
Leto [7]

Answer: 2.27\ m/s^2

Explanation:

Given

Length of the race track L=200\ m

the radius of curvature of the track r=29.5\ m

time taken to run on track is t=24.4\ s

Speed of runner is

\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s

Centripetal acceleration is

\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2

5 0
3 years ago
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