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3 years ago

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uniformly charged conducting sphere of 1.2 m diameter has surface charge density 8.1 mC/m2. Find (a) the net charge on the spher

e and (b) the total electric flux leaving the surface.

1 answer:

Stels [109]3 years ago

5 0

Answer:

(a) = 3.7 × 10⁻⁵

(b) = 4.1 × 10⁻⁶ N.M²/C

Explanation:

(a) Diameter of the sphere, d = 1.2 m

Radius of the sphere, r = 0.6 m

Surface charge density, = 8.1 mC/m2 = 8.1 × 10⁻⁶ C/m²

Total charge on the surface of the sphere,

Q = Charge density × Surface area

= 4πr²σ

= 4 (3.14) (0.12²) (8.1 × 10⁻⁶)

= 3.66 × 10⁻⁵C

≅ 3.7 × 10⁻⁵C

Therefore, the net charge on the sphere is 3.7 × 10⁻⁵C

(b)

Total electric flux (∅)

=Q / ε₀

ε₀ = 8.854 × 10⁻¹² N⁻¹C² m⁻²

Q = 3.66 × 10⁻⁵C

= 3.66 × 10⁻⁵ / 8.854 × 10⁻¹²

= 4.1 × 10⁻⁶ N.M²/C

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Rus_ich [418]

Answer:

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Explanation:

Given that

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As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

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$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=$

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Answer:

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Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.

Speed of the horse = 12.8 m/s

Distance = Speed × Time

⇒Distance = 12.8×0.67

⇒Distance = 8.576 m

Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.

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A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. Whe

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Answer:

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2 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

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3 years ago

A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere

Tanzania [10]

Answer:

The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

Explanation:

Given that,

Radius = 5 cm

Charge density $J= 400\ nC/m^3$

We need to calculate the total charge Q of the sphere

Using formula of charge

$q=\rho V$

Where, $\rho$ = charge density

V = volume

Put the value into the formula

$q=\rho\times(\dfrac{4}{3}\pi r^3)$

Put the value into the formula

$q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3$

$q=2.094\times10^{-10}\ C$

Hence, The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

6 0

2 years ago