Energy and Potential An electric field is expressed in rectangular coordinates by E = 6x²ax+6yay + 4az V/m Find: a. VN if points M and N are specified by M(2, 6,-1) and N(-3,-3, 2) b. VM if V=0 at Q(4,-2,-35) c. Vif V=2 at P(1, 2,-4).
Energy and Potential An electric field is expressed in rectangular coordinates by E = 6x²ax+6yay + 4az V/m Find: a. VN if points M and N are specified by M(2, 6,-1) and N(-3,-3, 2) b. VM if V=0 at Q(4,-2,-35) c. Vif V=2 at P(1, 2,-4).
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electromagnetic field)I want a detailed solution because my teacher changes the numbers. I want a detailed solution. Understand the solution
![Energy and Potential
• An electric field is expressed in rectangular coordinates by
E = 6x²ax+6yay + 4az V/m
• Find:
a. VMN if points M and N are specified by M(2, 6,-1) and N(-3,-3, 2)
b. VM if V=0 at Q(4,-2,-35)
c. Vif V 2 at P(1, 2,-4).
• Solution:
• Exercise (a):
• Ē.dī= (6x²ax + 6yay + 4az). (dxax + dyāy + dzāz)
• Ē. dL = 6x² dx + 6ydy + 4dz
• V = -√ E.dL
.
.
Exercise (b):
• E.dī= (6x¹ax + 6ya, +4āz). (dxax +
dya, + dzā₂)
• E.dL = 6x²dx + 6ydy + 4dz
• VMO = -26x² dx + 2₂6ydy + 4dz
-[2x³12-[3y²1₂-[42]-35
• VMO
• VMN = -√²₂ 6x² dx + 6ydy + f₂¹4dz
VMN-[2x³]23 - [3y2]63 - [4z]z¹
=
VMN-70-81 + 12
• VMN-139 V
●
●
VMO
.
.
VMO -120 V
.
=
- Exercise (c):
112-96-136
• E. dL = (6x²ax + 6yāy +4āz). (dxāx+
dya, + dzāz)
• E. dL = 6x²dx + 6ydy + 4dz
VNP −₁³6x²dx + √₂³ 6ydy + f²354dz
VNP-[2x³]3- [3y²123- [42]24
VNP = 56 – 15 – 24
VNP = 17 V
=
• VMO = VM-Vo
• VM-[16+ 108-4]-0
• VM = -120 V
• VNP =Vn−Vp
Vn =VNP+Vp
.
• VN=17+2
VN = 19 V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd8d3d6c-3dee-4c15-b900-608d94f19202%2F4d3da9be-c34e-42cc-9626-c672f3caa640%2F9ulj281_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Energy and Potential
• An electric field is expressed in rectangular coordinates by
E = 6x²ax+6yay + 4az V/m
• Find:
a. VMN if points M and N are specified by M(2, 6,-1) and N(-3,-3, 2)
b. VM if V=0 at Q(4,-2,-35)
c. Vif V 2 at P(1, 2,-4).
• Solution:
• Exercise (a):
• Ē.dī= (6x²ax + 6yay + 4az). (dxax + dyāy + dzāz)
• Ē. dL = 6x² dx + 6ydy + 4dz
• V = -√ E.dL
.
.
Exercise (b):
• E.dī= (6x¹ax + 6ya, +4āz). (dxax +
dya, + dzā₂)
• E.dL = 6x²dx + 6ydy + 4dz
• VMO = -26x² dx + 2₂6ydy + 4dz
-[2x³12-[3y²1₂-[42]-35
• VMO
• VMN = -√²₂ 6x² dx + 6ydy + f₂¹4dz
VMN-[2x³]23 - [3y2]63 - [4z]z¹
=
VMN-70-81 + 12
• VMN-139 V
●
●
VMO
.
.
VMO -120 V
.
=
- Exercise (c):
112-96-136
• E. dL = (6x²ax + 6yāy +4āz). (dxāx+
dya, + dzāz)
• E. dL = 6x²dx + 6ydy + 4dz
VNP −₁³6x²dx + √₂³ 6ydy + f²354dz
VNP-[2x³]3- [3y²123- [42]24
VNP = 56 – 15 – 24
VNP = 17 V
=
• VMO = VM-Vo
• VM-[16+ 108-4]-0
• VM = -120 V
• VNP =Vn−Vp
Vn =VNP+Vp
.
• VN=17+2
VN = 19 V
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