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3 years ago

Is electrostatic induction an attraction between positive and negative charges

Physics

1 answer:

Natasha_Volkova [10]3 years ago

8 0

Explanation:

Electrostatic induction, also known as "electrostatic influence" or simply "influence" in Europe and Latin America, is a redistribution of electric charge in an object, caused by the influence of nearby charges.[1] In the presence of a charged body, an insulated conductor develops a positive charge on one end and a negative charge on the other end.[1] Induction was discovered by British scientist John Canton in 1753 and Swedish professor Johan Carl Wilcke in 1762.[2] Electrostatic generators, such as the Wimshurst machine, the Van de Graaff generator and the electrophorus, use this principle. Due to induction, the electrostatic potential (voltage) is constant at any point throughout a conductor.[3] Electrostatic Induction is also responsible for the attraction of light nonconductive objects, such as balloons, paper or styrofoam scraps, to static electric charges. Electrostatic induction laws apply in dynamic situations as far as the quasistatic approximation is valid. Electrostatic induction should not be confused with Electromagnetic induction

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A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in

Rashid [163]

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J

But ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

8 0

3 years ago

Read 2 more answers

What is the logical oreder for these words : verse, interlude,verse,refrain, introduction,refrain,coda,refrain,verse

Nikolay [14]

Sorry to say but I know that t(e introduction is first and the coda is last

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4 years ago

A diffraction grating with 230 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At w

mr_godi [17]

Answer:

θ₁ = 5.4°

θ₂ = 10.86°

Explanation:

The angle ca be found by using grating equation:

mλ = d Sinθ

where,

m = order of diffraction

λ = wavelength = 405.3 nm = 4.053 x 10⁻⁷ m

d = grating element = 1/230 lines/mm = 0.0043 mm/line = 4.3 x 10⁻⁶ m/line

θ = angle = ?

FOR m = 1:

(1)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₁ = 0.09425

θ₁ = Sin⁻¹(0.09425)

θ₁ = 5.4°

FOR m = 2:

(2)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₂ = 0.1885

θ₂ = Sin⁻¹(0.1885)

θ₂ = 10.86°

6 0

3 years ago

Which of the following best describes seamounts and islands of the deep ocean basins?

Rina8888 [55]

Answer: The correct option is (d)

lava flows built up from the ocean floor by multiple, summit and flank eruptions

Explanation:

Piles of baseltic lava flows built up from the ocean floor by multiple summit and flank eruptions describes seamounts and islands of the deep ocean basins.

4 0

4 years ago

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

6 0

3 years ago