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3 years ago

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

$EMF = M \frac{di}{dt}$

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

$EMF = 0.1 V/s$

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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:

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

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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