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.

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!!!!

The correct answers are as follows:

(a) The equation for the x-component of the electric field is[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]

(b) The equation for the y-component of the electric field is [tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]

(c) The equation for the z-component of the electric field is [tex]\[ E_z = 5.32x^2y^4z \][/tex]

(d) the magnitude of the electric field at the point x13-(-5.0, 2.0, 1.5) m in units of newtons per coulomb is 14907.5 N/C.

To find the components of the electric field, we need to take the negative gradient of the electric potential function V(x,y,z). The gradient of a function is a vector field whose components are the partial derivatives of the function with respect to each variable. The electric field [tex]\( \vec{E} \)[/tex] is related to the electric potential [tex]\( V \)[/tex] by the equation:

[tex]\[ \vec{E} = -\nabla V \][/tex]

where [tex]\( \nabla \)[/tex] is the gradient operator.

(a) The x-component of the electric field [tex]\( E_x \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( x \)[/tex] :

[tex]\[ E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_x = -3α(2xy^3) + 2γ(2xy^4z^2) \][/tex]

[tex]\[ E_x = -6αxy^3 + 4γxy^4z^2 \][/tex]

Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :

[tex]\[ E_x = -6(2.5)xy^3 + 4(1.33)xy^4z^2 \][/tex]

[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]

(b) The y-component of the electric field [tex]\( E_y \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( y \)[/tex] :

[tex]\[ E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex] [tex]\[ E_y = -3αx^2(3y^2) + 2γx^2(4y^3z^2) \][/tex]

[tex]\[ E_y = -9αx^2y^2 + 8γx^2y^3z^2 \][/tex]

Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :

[tex]\[ E_y = -9(2.5)x^2y^2 + 8(1.33)x^2y^3z^2 \][/tex]

[tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]

(c) The z-component of the electric field

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( z \)[/tex] :

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_z = -2γx^2y^4(-2z) \][/tex]

[tex]\[ E_z = 4γx^2y^4z \][/tex]

Substituting the given value of[tex]\( γ \)[/tex] :

[tex]\[ E_z = 4(1.33)x^2y^4z \]\\ E_z = 5.32x^2y^4z \][/tex]

(d) To calculate the magnitude of the electric field at the point[tex]\( P(-5.0, 2.0, 1.5) \)[/tex] m, we first substitute [tex]\( x = -5.0 \)[/tex] m, [tex]\( y = 2.0 \)[/tex] m, and [tex]\( z = 1.5 \)[/tex] m into the equations for[tex]\( E_x \)[/tex], [tex]\( E_y \)[/tex] , and [tex]\( E_z \)[/tex] :

[tex]E_x = -15(-5.0)(2.0)^3 + 5.32(-5.0)(2.0)^4(1.5)^2 \]\\ E_x = 1500 - 5.32(-5.0)(16)(2.25) \]\\ E_x = 1500 - (-851.88) \]\\ E_x = 1500 + 851.88 \]\\ E_x = 2351.88 \text{ N/C} \][/tex]

[tex]\[ E_y = -22.5(-5.0)^2(2.0)^2 + 10.64(-5.0)^2(2.0)^3(1.5)^2 \]\\ E_y = -22.5(25)(4) + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 4752 \]\\ E_y = 2552 \text{ N/C} \][/tex]

[tex]\[ E_z = 5.32(-5.0)^2(2.0)^4(1.5) \]\\ E_z = 5.32(25)(16)(1.5) \]\\ E_z = 5.32(600) \]\\ E_z = 3192 \text{ N/C} \][/tex]

Now, the magnitude of the electric field \( \vec{E} \) is given by:

[tex]\[ |\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{(2351.88)^2 + (2552)^2 + (3192)^2} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{5527641.64 + 6512644 + 10180416} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{22220502.64} \][/tex]

[tex]\[ |\vec{E}| \approx 14907.5 \text{ N/C} \][/tex]

Therefore, the magnitude of the electric field at the point [tex]\( P(-5.0, 2.0, 1.5) \)[/tex]m is approximately [tex]\( 14907.5 \)[/tex] N/C.

Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highway, she knows that she must exit the highway within 55

min

min

if she is not going to arrive late. Her exit is 43

mi

miHow much time would it take at the posted 60 mph speed?

away.

How are the electric field lines around a positive charge affected when a second positive charge is rear it?


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