Tensio electrica – problemata et solutiones

1. Potentiale electricum in puncto sito 1 cm ab onere 5.0 μC determina . Constans Coulombiana (k) = 9 × 10⁶ ^Nm² C⁻² ^{, 1} μC = ^10⁻⁶ C.

Notum:

Distantia a carica (r) = 1 cm = 1/100 m = 0.01 m = ^10⁻² m

Carica (q) = 5.0 μ C = 5.0 × ^10⁻⁶ C

Constans Coulombiana ( ^k ) = 9 × ^10⁶ Nm² ^C⁻²

Quaesitum: Potentia electrica (V)

solution:

Potentia electrica:

Potentia electrica est 4.5 × ^10⁶ Voltia.

[irp]

2. Praecipe Q₁ = 5.0 μC et crimen Q₂ = 6.0 μC. Constans Coulombiana (k) = 9 × 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Punctum A inter onera locatum est.. Potentiale electricum in puncto A determina.

Notum:

Carica Q1 = -5.0 μC = -5.0 ^× _10⁻⁶C

Distantia puncti A a _Q1 = 10 cm = 0.1 m = ^10⁻¹ m

Carica Q² = 6.0 μ _C = 6.0 × ^10⁻⁶ C

Distantia puncti a a _Q² = 10 cm = 0.1 m = ^10⁻¹ m

Constans Coulombiana ( ^k ) = 9 × ^10⁶ Nm² ^C⁻²

Quaesitum: Potentia electrica ad punctum A

solution:

Potentia electrica 1 :

Potentia electrica 2 :

Potentia electrica ad punctum A :

V = _V² – _V₁

V = (54 – 45) × ^10⁴

V = 9 × ^10⁴

[irp]

3. Praecipe q₁ = 5.0 μC et crimen q₂ = 6.0 μC. Constans Coulombiana (k) = 9 × 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Potentiale electricum in puncto A determina.

Notum:

Carica Q1 = -5.0 μC = -5.0 ^× _10⁻⁶C

Distantia puncti A a _Q² = 40 cm = 0.4 m = 4 × ^10⁻¹ m

Carica Q² = 6.0 μ _C = 6.0 × ^10⁻⁶ C

Distantia puncti A a _Q² = 50 cm = 0.5 m = 5 × ^10⁻¹ m

Constans Coulombiana ( ^k ) = 9 × ^10⁶ Nm² ^C⁻²

Quaesitum: Potentia electrica ad punctum A

solution:

Potentia electrica 1 :

Potentia electrica 2 :

Potentia electrica ad punctum A :

_V = _V₁ + V₂

V = (-11.25 + 10.8) × ^10⁴

V = -0.45 × ^10³

V = -4.5 × ^10³

Potentia electrica in puncto A est -4.5 × ^10³ Voltia.