Un cañón de electrones dispara electrones (q = -e, me = 9.1 × 10 -31 kg) hacia una placa metálica que está a 4.0 mm de distancia
1 answer:
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Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ =
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| =
(|r₂₁|)² = 148.25
= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Answer:
C = 17 i^ - 7 j^ + 16 k^ , | C| = 24.37
Explanation:
To work the vactor component method, we add the sum in each axis
C = A + B = (Aₓ + Bₓ) i ^ + ( +
) i ^ + (
+
) k ^
Cₓ = 12+ 5 = 17
= -37 +30 = -7
= 58 -42 = 16
Resulting vector
C = 17 i ^ - 7j ^ + 16k ^
The mangitude of the vector is
| C | = √ c²
| C | = √( 17² + 7² + 16²)
| C| = 24.37