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Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

A bullet of mass 20g is a fired with a velocity of 50mls from a gun of mass 2.5kg. Calculate the recoil velocity of the gun.

jok3333 [9.3K]

Answer:

v = - 0.4 m/s

Explanation:

Initial momentum is zero as neither gun nor bullet have velocity

Conservation of momentum

0 = 0.020(50) + 2.5v

v = - 0.4 m/s

3 0

3 years ago

How temperature, surface tension and the diameter of the tube affect capillary action.

Alexus [3.1K]

The surface tension acts to hold the surface intact. Capillary action occurs when the adhesion to the surface material is stronger than the cohesive forces between the water molecules. ... Water wants to stick to the glass and surface tension will push the water up, until the force of gravity prevents further rise.

Narrower tube openings allow capillary action to pull water higher

5 0

3 years ago

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.500 m2 . At

Anastaziya [24]

Answer:

31.8 × 10⁻⁴ J = 3.18 mJ

Explanation:

We know the intensity I of a wave is I = P/A where P = power and A = area = 0.500 m²

The intensity of an electromagnetic wave is also equal to I = E₀²/μ₀c

where E₀ = maximum electric field strength = √2E where E = rms value of electric field = 0.0200 N/C, μ₀ = 4π × 10⁻⁷ H/m ,c = 3 × 10⁸ m/s

P/A = E₀²/μ₀c = 2E²/μ₀c

P = 2E²A/μ₀c = 2 × (0.02 N/C)² × 0.5 m²/(4π × 10⁻⁷ H/m × 3 × 10⁸ m/s)

= 1.06 × 10⁻⁴ W = 0.106 mW

Since P = E/t where E = Energy and t = time

E = Pt with t = 30 s

E = 1.06 × 10⁻⁴ W × 30 s = 31.8 × 10⁻⁴ J = 3.18 mJ

So the wave carries 3.18 mJ of energy through the window in 30 s

5 0

3 years ago

What is the density of lead (in g/cm3) if a rectangular bar measuring 0.50 cm in height, 1.55 cm in width, and 25.00 cm in lengt

Otrada [13]

If you look at the units of density, g/cm^3 you will see it's grams divided by volume (cm^3).

218.9 g / (0.5*1.55*25 cm^3) = 11.3 g/cm^3

4 0

4 years ago

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How can I convert this?Please answer with solution. Thank you.​

How can I convert this?Please answer with solution. Thank you.