(b) A potential is given by the expression $V=2(x+1)^2(y+2)^2(z+3)^2$ volts in free space. Determine each of the following at point $P (1,2,3)$: (i) potential; (ii) electric field; (iii) electric flux density.
Answer (1)
Calculate the potential V at point P ( 1 , 2 , 3 ) by substituting the coordinates into the given potential function, resulting in V = 4608 volts.
Determine the electric field E by finding the negative gradient of the potential V and evaluating it at point P , which gives E = − 4608 a x − 2304 a y − 1536 a z V/m.
Compute the electric flux density D using the relation D = ϵ 0 E , where ϵ 0 is the permittivity of free space, yielding D = − 4.08 × 1 0 − 8 a x − 2.04 × 1 0 − 8 a y − 1.36 × 1 0 − 8 a z C/m 2 .
State the final answers for potential, electric field, and electric flux density at point P: V = 4608 V , E = − 4608 a x − 2304 a y − 1536 a z V/m , and D = − 4.08 × 1 0 − 8 a x − 2.04 × 1 0 − 8 a y − 1.36 × 1 0 − 8 a z C/m 2 .
Explanation
Problem Setup We are given the potential function V = 2 ( x + 1 ) 2 ( y + 2 ) 2 ( z + 3 ) 2 and asked to find the potential, electric field, and electric flux density at point P ( 1 , 2 , 3 ) .
Calculating Potential (i) To find the potential at P ( 1 , 2 , 3 ) , we substitute x = 1 , y = 2 , and z = 3 into the potential function:
V = 2 ( 1 + 1 ) 2 ( 2 + 2 ) 2 ( 3 + 3 ) 2 = 2 ( 2 ) 2 ( 4 ) 2 ( 6 ) 2 = 2 ( 4 ) ( 16 ) ( 36 ) = 4608
So, the potential at point P is 4608 volts.
Calculating Electric Field (ii) To find the electric field E , we need to calculate the negative gradient of the potential V :
E = − ∇ V = − ( ∂ x ∂ V a x + ∂ y ∂ V a y + ∂ z ∂ V a z ) First, we find the partial derivatives: ∂ x ∂ V = 4 ( x + 1 ) ( y + 2 ) 2 ( z + 3 ) 2 ∂ y ∂ V = 4 ( x + 1 ) 2 ( y + 2 ) ( z + 3 ) 2 ∂ z ∂ V = 4 ( x + 1 ) 2 ( y + 2 ) 2 ( z + 3 ) Now, we evaluate these partial derivatives at point P ( 1 , 2 , 3 ) :
∂ x ∂ V ∣ P = 4 ( 1 + 1 ) ( 2 + 2 ) 2 ( 3 + 3 ) 2 = 4 ( 2 ) ( 16 ) ( 36 ) = 4608 ∂ y ∂ V ∣ P = 4 ( 1 + 1 ) 2 ( 2 + 2 ) ( 3 + 3 ) 2 = 4 ( 4 ) ( 4 ) ( 36 ) = 2304 ∂ z ∂ V ∣ P = 4 ( 1 + 1 ) 2 ( 2 + 2 ) 2 ( 3 + 3 ) = 4 ( 4 ) ( 16 ) ( 6 ) = 1536 Thus, the electric field at point P is: E = − 4608 a x − 2304 a y − 1536 a z V/m
Calculating Electric Flux Density (iii) To find the electric flux density D , we use the relationship D = ϵ 0 E , where ϵ 0 is the permittivity of free space, approximately 8.854 × 1 0 − 12 F/m. D = ϵ 0 E = ϵ 0 ( − 4608 a x − 2304 a y − 1536 a z ) D = − 4608 ϵ 0 a x − 2304 ϵ 0 a y − 1536 ϵ 0 a z D = − 4608 ( 8.854 × 1 0 − 12 ) a x − 2304 ( 8.854 × 1 0 − 12 ) a y − 1536 ( 8.854 × 1 0 − 12 ) a z D = − 4.0799 × 1 0 − 8 a x − 2.03996 × 1 0 − 8 a y − 1.35997 × 1 0 − 8 a z C/m 2 So, the electric flux density at point P is approximately: D = − 4.08 × 1 0 − 8 a x − 2.04 × 1 0 − 8 a y − 1.36 × 1 0 − 8 a z C/m 2
Final Answer In summary, at point P ( 1 , 2 , 3 ) :
(i) The potential is V = 4608 volts. (ii) The electric field is E = − 4608 a x − 2304 a y − 1536 a z V/m. (iii) The electric flux density is D = − 4.08 × 1 0 − 8 a x − 2.04 × 1 0 − 8 a y − 1.36 × 1 0 − 8 a z C/m 2 .
Examples
Understanding potential, electric field, and electric flux density is crucial in various applications, such as designing electronic devices and ensuring their proper functioning. For instance, when designing a capacitor, engineers need to calculate the electric field and flux density between the capacitor plates to determine its capacitance and voltage rating. Similarly, in designing high-voltage power lines, it's essential to calculate the electric field around the conductors to prevent electrical breakdown and ensure safety. These calculations help in optimizing the design and performance of electrical and electronic systems.
