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dedylja
1 month ago
5

The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym

+Czn+DV=Axl+Bym+Czn+Dwhere AA, BB, CC, DD, ll, mm, and nn are constants. The units of AA, BB, CC, and DD are such that if xx, yy, and zz are in meters, then VV is in volts. You measure VV and each component of the electric field at four points and obtain these results:Point (x,y,z)(m) V(V) Ex(V/m) Ey(V/m) Ez(V/m) 1 (0, 0, 0) 10.0 0 0 0 2 (1.00, 0, 0) 4.0 16.0 0 0 3 (0, 1.00, 0) 6.0 0 16.0 0 4 (0, 0, 1.00) 8.0 0 0 16.01. Use the data to calculate A.2. Use the data to calculate B3. Use the data to calculate C4. Use the data to calculate D5. Use the data to calculate E6. Use the data to calculate l7. Use the data to calculate m8. Use the data to calculate n
Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
4 0

Answer:

Considering the electric potential, V = Ax^l+By^m+Cz^n+D

The electric field components are:

E_x = \frac{-dV}{dx} = -Alx^l^-^1

E_y = \frac{-dV}{dy} = - Bmy^m^-^1

E_z = \frac{-dV}{dz} = - nCzn^n^-^1

Let's find the potential differences for the specified points.

V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10

=> D = 10

V(1, 0, 0) = 4V => A + 10 = 4

For point A, we deduce:

A = 4 - 10

A = -6

V(0, 1, 0) = 6V => B + 10 = 6

For point B, we conclude:

B = 6 - 10

B = -4

V(0, 0, 1) = 8V => C + 10 = 4

For point C, we establish:

C = 8 - 10

C = -2

Now, to calculate the electric field intensity at the given locations, we perform the following steps:

E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16

Al = -16

For l, the calculation yields:

l = \frac{-16}{A}

From the prior findings, we have A = -6

l = \frac{-16}{-6}

l = \frac{8}{3}

E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16

Bm = -16

Next, for m, the solution is:

m = \frac{-16}{A}

Thus, we find B = -4

m = \frac{-16}{-4}

m = 4

E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16

nC = - 16

Finally, for n, the determination results in:

n = \frac{-16}{C}

Hence, C = -2

n = \frac{-16}{-2}

n = 8

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