(b) A potential is given by the expression [tex]$V=2(x+1)^2(y+2)^2(z+3)^2$[/tex] volts in free space. Determine each of the following at point [tex]$P (1,2,3)$[/tex]:
(i) potential;
(ii) electric field;
(iii) electric flux density.

Calculate the potential V at P ( 1 , 2 , 3 ) using the given formula: V = 2 ( x + 1 ) 2 ( y + 2 ) 2 ( z + 3 ) 2 , resulting in V = 4608 volts.
Determine the electric field E by finding the negative gradient of V and evaluating it at P ( 1 , 2 , 3 ) , which gives E = − 4608 a x ​ − 2304 a y ​ − 1536 a z ​ V/m.
Compute the electric flux density D using D = ϵ 0 ​ E , where ϵ 0 ​ = 8.854 × 1 0 − 12 F/m, yielding D = − 4.0799 × 1 0 − 8 a x ​ − 2.03996 × 1 0 − 8 a y ​ − 1.35997 × 1 0 − 8 C/m 2 .
State the final answers for potential, electric field, and electric flux density at point P ( 1 , 2 , 3 ) . V = 4608 V , E = − 4608 a x ​ − 2304 a y ​ − 1536 a z ​ V/m , 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 ​

Explanation

Problem Setup We are given the potential function V = 2 ( x + 1 ) 2 ( y + 2 ) 2 ( z + 3 ) 2 volts and asked to find the potential, electric field, and electric flux density at point P ( 1 , 2 , 3 ) . We'll tackle each part step by step.

Calculating Potential First, we calculate the potential V at point P ( 1 , 2 , 3 ) by substituting the coordinates into the potential function:


V ( 1 , 2 , 3 ) = 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 Next, we calculate the electric field E by taking the negative gradient of the potential: E = − ∇ V = − ( ∂ x ∂ V ​ a x ​ + ∂ y ∂ V ​ a y ​ + ∂ z ∂ V ​ a z ​ ) .

We need to compute 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 ​ ( 1 , 2 , 3 ) = 4 ( 1 + 1 ) ( 2 + 2 ) 2 ( 3 + 3 ) 2 = 4 × 2 × 16 × 36 = 4608 ∂ y ∂ V ​ ( 1 , 2 , 3 ) = 4 ( 1 + 1 ) 2 ( 2 + 2 ) ( 3 + 3 ) 2 = 4 × 4 × 4 × 36 = 2304 ∂ z ∂ V ​ ( 1 , 2 , 3 ) = 4 ( 1 + 1 ) 2 ( 2 + 2 ) 2 ( 3 + 3 ) = 4 × 4 × 16 × 6 = 1536
Thus, the electric field at point P ( 1 , 2 , 3 ) is:
E ( 1 , 2 , 3 ) = − ( 4608 a x ​ + 2304 a y ​ + 1536 a z ​ )

Calculating Electric Flux Density Finally, we calculate the electric flux density D using the relation D = ϵ 0 ​ E , where ϵ 0 ​ is the permittivity of free space ( ϵ 0 ​ ≈ 8.854 × 1 0 − 12 F/m).

So, at point P ( 1 , 2 , 3 ) :
D ( 1 , 2 , 3 ) = ϵ 0 ​ E ( 1 , 2 , 3 ) = 8.854 × 1 0 − 12 × ( − 4608 a x ​ − 2304 a y ​ − 1536 a z ​ )
D ( 1 , 2 , 3 ) = ( − 4.0799 × 1 0 − 8 a x ​ − 2.03996 × 1 0 − 8 a y ​ − 1.35997 × 1 0 − 8 a z ​ ) C/m 2

Final Answer In summary, at point P ( 1 , 2 , 3 ) :

(i) The potential V = 4608 volts. (ii) The electric field E = − 4608 a x ​ − 2304 a y ​ − 1536 a z ​ V/m. (iii) The electric flux density 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 .
Examples
Understanding potential, electric field, and electric flux density is crucial in designing and analyzing electrical systems. For instance, when designing a capacitor, engineers need to calculate the electric field between the plates to determine the capacitor's capacitance and voltage rating. Similarly, in designing high-voltage power lines, it's essential to calculate the electric field around the conductors to ensure safety and prevent electrical breakdown of the air. These calculations help in optimizing the design and ensuring the reliable operation of electrical devices and systems.

Answered by GinnyAnswer | 2025-07-08

At point P ( 1 , 2 , 3 ) , the potential is 4608 volts, the electric field is given by E = − 4608 a x ​ − 2304 a y ​ − 1536 a z ​ V/m, and the electric flux density is 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 .
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Answered by Anonymous | 2025-07-23