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3 years ago

10

An unknown charge sits on a conducting solid sphere of radius 9.0 cm. If the electric field 14 cm from the center of the sphere

has magnitude 2.7 103 N/C and is directed radially inward, what is the net charge on the sphere?

2 answers:

romanna [79]3 years ago

7 0

Answer: The net charge on the sphere is -5.9nC

Explanation:

The electric field outside the conducting solid sphere is given as:

E = (kq)/r^2

Here, q is the net charge on the sphere and r is the distance from the center of the sphere.

The net charge is calculated as follows;

q = Er^2/k

E = 2.7*10^3N/C

r = 14cm = 0.14m

k = 8.99*10^9Nm^2/C^2

q = (2.7*10^3)(0.14^2)/(8.99*10^9)

q = 5.9*10^-9C

= 5.9nC

As the electric field directed radially inward, the net charge on the sphere is negative.

Hence the net charge on the sphere is -5.9nC

spin [16.1K]3 years ago

7 0

Answer:

The net charge on the sphere is - 5.9 × 10⁻⁹ C = - 5.9 nC

Explanation:

The electric field (E) produced by a charge of magnitude Q, at a point with distance r away from the charge, is given by

E = kQ/r²

where k = Coulomb's constant = 9.0 x 10⁹ Nm²/C²

The electric field 14 cm (0.14 m) from the sphere is 2.7 × 10³ N/C

2.7 × 10³ = 9.0 x 10⁹ × Q/(0.14²)

Q = (2.7 × 10³ × 0.14²)/(9.0 x 10⁹) = 5.9 × 10⁻⁹ C

Since the charge is directed inwards towards the centre of the sphere, the sign on it will be -ve. Q = - 5.9 × 10⁻⁹ C

b) The electric field 9 cm (0.09 m) from the sphere will be

E = kQ/(0.09²) = (9 × 10⁹ × 5.9 × 10⁻⁹)/(0.09²) = 6555.56 = 6.56 × 10³ N/C

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Read 2 more answers

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

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Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

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The electric force exerted on a charge by an electric field is given by:

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