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topjm [15]
3 years ago
11

The electric potential in a certain region is given by the equation V(x,y,z) = 3αx2y3 - 2γx2y4z2 where the potential is in volts

when the positions are given in meters. The constants in this equation are α = 2.5 V/m5 and γ = 1.33 V/m8
(a) Write an equation for the x-component of the electric field.

(b) Write an equation for the y-component of the electric field.

(c) Write an equation for the z-component of the electric field.

(d) Calculate the magnitude of the electric field at the point x13-(-5.0, 2.0, 1.5) m in units of newtons per coulomb.
Physics
1 answer:
Mila [183]3 years ago
7 0

Answer:

a) An equation for the x-component of the electric field.

Eₓ = (-15xy³ + 5.32xy⁴z²) N/C

b) An equation for the y-component of the electric field.

Eᵧ = (-22.5x²y² + 10.64x²y³z²) N/C

c) An equation for the z-component of the electric field.

Ez = (5.32x²y⁴z) N/C

d) At (-5.0, 2.0, 1.5) m, the electric field is given as

E = (-357.6î + 2,538ĵ + 3,192ķ) N/C

Magnitude of the electric field = 4,093.7 N/C

Explanation:

The electric field is given by the negative of the gradient of the electric potential,

E = −grad V

E = - ∇V

The electric potential is given as

V(x,y,z) = 3αx²y³ - 2γx²y⁴z²

α = 2.5 V/m⁵ and γ = 1.33 V/m⁸

V(x,y,z) = 7.5x²y³ - 2.66x²y⁴z²

grad = ∇ = (∂/∂x)î + (∂/∂y)ĵ + (∂/∂z)ķ

E = -grad V = -∇V

= -[(∂V/∂x)î + (∂V/∂y)ĵ + (∂V/∂z)ķ

E = -(∂V/∂x)î - (∂V/∂y)ĵ - (∂V/∂z)ķ

E = Eₓî + Eᵧĵ + Ez ķ

a) An equation for the x-component of the electric field.

Eₓ = -(∂V/∂x) = -(∂/∂x)(V)

= -(∂/∂x)(7.5x²y³ - 2.66x²y⁴z²)

= -(15xy³ - 5.32xy⁴z²)

= (-15xy³ + 5.32xy⁴z²)

b) An equation for the y-component of the electric field.

Eᵧ = -(∂V/∂y) = -(∂/∂x)(V)

= -(∂/∂y)(7.5x²y³ - 2.66x²y⁴z²)

= -(22.5x²y² - 10.64x²y³z²)

= (-22.5x²y² + 10.64x²y³z²)

c) An equation for the z-component of the electric field.

Ez = -(∂V/∂z) = -(∂/∂x)(V)

= -(∂/∂z)(7.5x²y³ - 2.66x²y⁴z²)

= -(0 - 5.32x²y⁴z)

= (5.32x²y⁴z)

d) E = Eₓî + Eᵧĵ + Ez ķ

E = (-15xy³ + 5.32xy⁴z²)î + (-22.5x²y² + 10.64x²y³z²)ĵ + (5.32x²y⁴z) ķ

At (-5.0, 2.0, 1.5) m

x = -5 m

y = 2 m

z = 1.5 m

Eₓ = (-15xy³ + 5.32xy⁴z²)

= (-15×-5×2³) + (5.32×-5×2⁴×1.5²)

= 600 - 957.6 = -357.6

Eᵧ = (-22.5x²y² + 10.64x²y³z²)

= (-22.5×(-5)²×2²) + (10.64×(-5)²×2³×1.5²)

= -2250 + 4788 = 2538

Ez = (5.32x²y⁴z) = (5.32×(-5)²×2⁴×1.5)

= 3192

E = -357.6î + 2,538ĵ + 3,192ķ

Magnitude = /E/ = √[(-357.6)² + 2538² + 3192²]

= 4,093.6763135353 = 4,093.7 N/C

Hope this Helps!!!!

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The change in the potential energy of the  launched from a height into the pool without friction from the given height h is calculated by applying the following kinematic equation.

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A 50 W light bulb is plugged into a standard
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8 0
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. Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connecte
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Given :

Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connected with wires as shown in the circuit given in figure. The current flowing in both the circuits is the same.

To Find :

Will the heat produced in both the cases be equal.

Solution :

Heat released is given by :

H = i²Rt

Here, R is resistance and is given by :

R = \dfrac{\rho L}{A}

So,

H = i^2\times \dfrac{\rho L}{A} t\\\\H = \dfrac{i^2\rho Lt}{A}

Now, in the question every thing is constant except for the length of the wire and from above equation heat is directly proportional to the length of the wire.

So, heat produced by Reem's wire is more than Nain one.

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3 years ago
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Answer:

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Given,

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Assuming the intensity of sunlight(S) be equal to 1.1 x 10³ W/m².

Electromagnetic energy = ?

E = \mu V

E = (\dfrac{S}{c})\times V

where c is the speed of light.

E = (\dfrac{1.1\times 10^3}{3\times 10^8})\times 5.20

E = 1.907\times 10^{-5}\ J

Hence, Electromagnetic energy is equal to 1.907 x 10⁻⁵ J.

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