Datos:
q 1 = q 2 = q = + 9 nC = 9 ⋅ 10 − 9 C q_1 = q_2 = q = +9 \text{ nC} = 9 \cdot 10^{-9} \text{ C} q 1 = q 2 = q = + 9 nC = 9 ⋅ 1 0 − 9 C r ⃗ 1 = ( − 6 , 0 ) mm = ( − 6 ⋅ 10 − 3 , 0 ) m \vec{r}_1 = (-6, 0) \text{ mm} = (-6 \cdot 10^{-3}, 0) \text{ m} r 1 = ( − 6 , 0 ) mm = ( − 6 ⋅ 1 0 − 3 , 0 ) m r ⃗ 2 = ( 6 , 0 ) mm = ( 6 ⋅ 10 − 3 , 0 ) m \vec{r}_2 = (6, 0) \text{ mm} = (6 \cdot 10^{-3}, 0) \text{ m} r 2 = ( 6 , 0 ) mm = ( 6 ⋅ 1 0 − 3 , 0 ) m K = 9 ⋅ 10 9 N ⋅ m 2 ⋅ C − 2 K = 9 \cdot 10^9 \text{ N} \cdot \text{m}^2 \cdot \text{C}^{-2} K = 9 ⋅ 1 0 9 N ⋅ m 2 ⋅ C − 2 a) El origen de coordenadas ( 0 , 0 ) m (0, 0) \text{ m} ( 0 , 0 ) m El potencial eléctrico en un punto debido a una carga puntual se calcula con la fórmula V = K q r V = K \frac{q}{r} V = K r q . Para un sistema de cargas, el potencial total es la suma algebraica de los potenciales individuales.
V t o t a l = ∑ V i = V 1 + V 2 V_{total} = \sum V_i = V_1 + V_2 V t o t a l = ∑ V i = V 1 + V 2 La distancia de ambas cargas al origen es la misma:
r 1 = ∣ r ⃗ 1 ∣ = 6 ⋅ 10 − 3 m r_1 = |\vec{r}_1| = 6 \cdot 10^{-3} \text{ m} r 1 = ∣ r 1 ∣ = 6 ⋅ 1 0 − 3 m r 2 = ∣ r ⃗ 2 ∣ = 6 ⋅ 10 − 3 m r_2 = |\vec{r}_2| = 6 \cdot 10^{-3} \text{ m} r 2 = ∣ r 2 ∣ = 6 ⋅ 1 0 − 3 m V O = K q 1 r 1 + K q 2 r 2 = 2 K q r V_O = K \frac{q_1}{r_1} + K \frac{q_2}{r_2} = 2 K \frac{q}{r} V O = K r 1 q 1 + K r 2 q 2 = 2 K r q V O = 2 ⋅ ( 9 ⋅ 10 9 N ⋅ m 2 ⋅ C − 2 ) ⋅ 9 ⋅ 10 − 9 C 6 ⋅ 10 − 3 m V_O = 2 \cdot (9 \cdot 10^9 \text{ N} \cdot \text{m}^2 \cdot \text{C}^{-2}) \cdot \frac{9 \cdot 10^{-9} \text{ C}}{6 \cdot 10^{-3} \text{ m}} V O = 2 ⋅ ( 9 ⋅ 1 0 9 N ⋅ m 2 ⋅ C − 2 ) ⋅ 6 ⋅ 1 0 − 3 m 9 ⋅ 1 0 − 9 C V O = 2 ⋅ 81 6 ⋅ 10 3 V = 2 ⋅ 13.5 ⋅ 10 3 V = 27000 V V_O = 2 \cdot \frac{81}{6} \cdot 10^3 \text{ V} = 2 \cdot 13.5 \cdot 10^3 \text{ V} = 27000 \text{ V} V O = 2 ⋅ 6 81 ⋅ 1 0 3 V = 2 ⋅ 13.5 ⋅ 1 0 3 V = 27000 V El campo eléctrico en un punto debido a una carga puntual se calcula con la fórmula E ⃗ = K q r 2 r ^ \vec{E} = K \frac{q}{r^2} \hat{r} E = K r 2 q r ^ , donde r ^ \hat{r} r ^ es el vector unitario que apunta desde la carga hacia el punto de interés. Para un sistema de cargas, el campo eléctrico total es la suma vectorial de los campos individuales.
E ⃗ t o t a l = ∑ E ⃗ i = E ⃗ 1 + E ⃗ 2 \vec{E}_{total} = \sum \vec{E}_i = \vec{E}_1 + \vec{E}_2 E t o t a l = ∑ E i = E 1 + E 2 Para q 1 q_1 q 1 en ( − 6 ⋅ 10 − 3 , 0 ) (-6 \cdot 10^{-3}, 0) ( − 6 ⋅ 1 0 − 3 , 0 ) y el punto de interés en ( 0 , 0 ) (0,0) ( 0 , 0 ) :
r ⃗ 1 O = ( 0 − ( − 6 ⋅ 10 − 3 ) ) i ⃗ = 6 ⋅ 10 − 3 i ⃗ m \vec{r}_{1O} = (0 - (-6 \cdot 10^{-3})) \vec{i} = 6 \cdot 10^{-3} \vec{i} \text{ m} r 1 O = ( 0 − ( − 6 ⋅ 1 0 − 3 )) i = 6 ⋅ 1 0 − 3 i m r ^ 1 O = i ⃗ \hat{r}_{1O} = \vec{i} r ^ 1 O = i E ⃗ 1 = K q 1 r 1 2 r ^ 1 O = ( 9 ⋅ 10 9 ) 9 ⋅ 10 − 9 ( 6 ⋅ 10 − 3 ) 2 i ⃗ N/C \vec{E}_1 = K \frac{q_1}{r_1^2} \hat{r}_{1O} = (9 \cdot 10^9) \frac{9 \cdot 10^{-9}}{(6 \cdot 10^{-3})^2} \vec{i} \text{ N/C} E 1 = K r 1 2 q 1 r ^ 1 O = ( 9 ⋅ 1 0 9 ) ( 6 ⋅ 1 0 − 3 ) 2 9 ⋅ 1 0 − 9 i N/C E ⃗ 1 = 81 36 ⋅ 10 − 6 i ⃗ = 2.25 ⋅ 10 6 i ⃗ N/C \vec{E}_1 = \frac{81}{36 \cdot 10^{-6}} \vec{i} = 2.25 \cdot 10^6 \vec{i} \text{ N/C} E 1 = 36 ⋅ 1 0 − 6 81 i = 2.25 ⋅ 1 0 6 i N/C Para q 2 q_2 q 2 en ( 6 ⋅ 10 − 3 , 0 ) (6 \cdot 10^{-3}, 0) ( 6 ⋅ 1 0 − 3 , 0 ) y el punto de interés en ( 0 , 0 ) (0,0) ( 0 , 0 ) :
r ⃗ 2 O = ( 0 − 6 ⋅ 10 − 3 ) i ⃗ = − 6 ⋅ 10 − 3 i ⃗ m \vec{r}_{2O} = (0 - 6 \cdot 10^{-3}) \vec{i} = -6 \cdot 10^{-3} \vec{i} \text{ m} r 2 O = ( 0 − 6 ⋅ 1 0 − 3 ) i = − 6 ⋅ 1 0 − 3 i m r ^ 2 O = − i ⃗ \hat{r}_{2O} = -\vec{i} r ^ 2 O = − i E ⃗ 2 = K q 2 r 2 2 r ^ 2 O = ( 9 ⋅ 10 9 ) 9 ⋅ 10 − 9 ( 6 ⋅ 10 − 3 ) 2 ( − i ⃗ ) N/C \vec{E}_2 = K \frac{q_2}{r_2^2} \hat{r}_{2O} = (9 \cdot 10^9) \frac{9 \cdot 10^{-9}}{(6 \cdot 10^{-3})^2} (-\vec{i}) \text{ N/C} E 2 = K r 2 2 q 2 r ^ 2 O = ( 9 ⋅ 1 0 9 ) ( 6 ⋅ 1 0 − 3 ) 2 9 ⋅ 1 0 − 9 ( − i ) N/C E ⃗ 2 = − 2.25 ⋅ 10 6 i ⃗ N/C \vec{E}_2 = -2.25 \cdot 10^6 \vec{i} \text{ N/C} E 2 = − 2.25 ⋅ 1 0 6 i N/C El campo eléctrico total en el origen es:
E ⃗ O = E ⃗ 1 + E ⃗ 2 = ( 2.25 ⋅ 10 6 i ⃗ ) + ( − 2.25 ⋅ 10 6 i ⃗ ) = 0 ⃗ N/C \vec{E}_O = \vec{E}_1 + \vec{E}_2 = (2.25 \cdot 10^6 \vec{i}) + (-2.25 \cdot 10^6 \vec{i}) = \vec{0} \text{ N/C} E O = E 1 + E 2 = ( 2.25 ⋅ 1 0 6 i ) + ( − 2.25 ⋅ 1 0 6 i ) = 0 N/C b) El punto ( 0 , 3 ) mm (0, 3) \text{ mm} ( 0 , 3 ) mm Las cargas q 1 q_1 q 1 y q 2 q_2 q 2 están en ( − 6 , 0 ) (-6,0) ( − 6 , 0 ) y ( 6 , 0 ) (6,0) ( 6 , 0 ) mm, respectivamente. El punto de interés P P P es ( 0 , 3 ) (0,3) ( 0 , 3 ) mm. La distancia de ambas cargas al punto P es la misma, por simetría:
r P = ( ( 0 ) − ( − 6 ⋅ 10 − 3 ) ) 2 + ( ( 3 ⋅ 10 − 3 ) − 0 ) 2 = ( 6 ⋅ 10 − 3 ) 2 + ( 3 ⋅ 10 − 3 ) 2 m r_P = \sqrt{((0) - (-6 \cdot 10^{-3}))^2 + ((3 \cdot 10^{-3}) - 0)^2} = \sqrt{(6 \cdot 10^{-3})^2 + (3 \cdot 10^{-3})^2} \text{ m} r P = (( 0 ) − ( − 6 ⋅ 1 0 − 3 ) ) 2 + (( 3 ⋅ 1 0 − 3 ) − 0 ) 2 = ( 6 ⋅ 1 0 − 3 ) 2 + ( 3 ⋅ 1 0 − 3 ) 2 m r P = 36 ⋅ 10 − 6 + 9 ⋅ 10 − 6 = 45 ⋅ 10 − 6 = 45 ⋅ 10 − 3 m r_P = \sqrt{36 \cdot 10^{-6} + 9 \cdot 10^{-6}} = \sqrt{45 \cdot 10^{-6}} = \sqrt{45} \cdot 10^{-3} \text{ m} r P = 36 ⋅ 1 0 − 6 + 9 ⋅ 1 0 − 6 = 45 ⋅ 1 0 − 6 = 45 ⋅ 1 0 − 3 m El potencial eléctrico en el punto P es:
V P = K q 1 r P + K q 2 r P = 2 K q r P V_P = K \frac{q_1}{r_P} + K \frac{q_2}{r_P} = 2 K \frac{q}{r_P} V P = K r P q 1 + K r P q 2 = 2 K r P q V P = 2 ⋅ ( 9 ⋅ 10 9 N ⋅ m 2 ⋅ C − 2 ) ⋅ 9 ⋅ 10 − 9 C 45 ⋅ 10 − 3 m V_P = 2 \cdot (9 \cdot 10^9 \text{ N} \cdot \text{m}^2 \cdot \text{C}^{-2}) \cdot \frac{9 \cdot 10^{-9} \text{ C}}{\sqrt{45} \cdot 10^{-3} \text{ m}} V P = 2 ⋅ ( 9 ⋅ 1 0 9 N ⋅ m 2 ⋅ C − 2 ) ⋅ 45 ⋅ 1 0 − 3 m 9 ⋅ 1 0 − 9 C V P = 162 45 ⋅ 10 3 V ≈ 24150 V V_P = \frac{162}{\sqrt{45}} \cdot 10^3 \text{ V} \approx 24150 \text{ V} V P = 45 162 ⋅ 1 0 3 V ≈ 24150 V Para el campo eléctrico, necesitamos los vectores unitarios.
Vector desde q 1 q_1 q 1 a P:
r ⃗ 1 P = ( 0 − ( − 6 ⋅ 10 − 3 ) ) i ⃗ + ( 3 ⋅ 10 − 3 − 0 ) j ⃗ = ( 6 ⋅ 10 − 3 i ⃗ + 3 ⋅ 10 − 3 j ⃗ ) m \vec{r}_{1P} = (0 - (-6 \cdot 10^{-3})) \vec{i} + (3 \cdot 10^{-3} - 0) \vec{j} = (6 \cdot 10^{-3} \vec{i} + 3 \cdot 10^{-3} \vec{j}) \text{ m} r 1 P = ( 0 − ( − 6 ⋅ 1 0 − 3 )) i + ( 3 ⋅ 1 0 − 3 − 0 ) j = ( 6 ⋅ 1 0 − 3 i + 3 ⋅ 1 0 − 3 j ) m r ^ 1 P = 6 ⋅ 10 − 3 i ⃗ + 3 ⋅ 10 − 3 j ⃗ 45 ⋅ 10 − 3 = 6 45 i ⃗ + 3 45 j ⃗ \hat{r}_{1P} = \frac{6 \cdot 10^{-3} \vec{i} + 3 \cdot 10^{-3} \vec{j}}{\sqrt{45} \cdot 10^{-3}} = \frac{6}{\sqrt{45}} \vec{i} + \frac{3}{\sqrt{45}} \vec{j} r ^ 1 P = 45 ⋅ 1 0 − 3 6 ⋅ 1 0 − 3 i + 3 ⋅ 1 0 − 3 j = 45 6 i + 45 3 j E ⃗ 1 = K q 1 r P 2 r ^ 1 P = ( 9 ⋅ 10 9 ) 9 ⋅ 10 − 9 ( 45 ⋅ 10 − 3 ) 2 ( 6 45 i ⃗ + 3 45 j ⃗ ) N/C \vec{E}_1 = K \frac{q_1}{r_P^2} \hat{r}_{1P} = (9 \cdot 10^9) \frac{9 \cdot 10^{-9}}{(\sqrt{45} \cdot 10^{-3})^2} \left( \frac{6}{\sqrt{45}} \vec{i} + \frac{3}{\sqrt{45}} \vec{j} \right) \text{ N/C} E 1 = K r P 2 q 1 r ^ 1 P = ( 9 ⋅ 1 0 9 ) ( 45 ⋅ 1 0 − 3 ) 2 9 ⋅ 1 0 − 9 ( 45 6 i + 45 3 j ) N/C E ⃗ 1 = 81 45 ⋅ 10 − 6 ( 6 45 i ⃗ + 3 45 j ⃗ ) = 1.8 ⋅ 10 6 ( 6 45 i ⃗ + 3 45 j ⃗ ) N/C \vec{E}_1 = \frac{81}{45 \cdot 10^{-6}} \left( \frac{6}{\sqrt{45}} \vec{i} + \frac{3}{\sqrt{45}} \vec{j} \right) = 1.8 \cdot 10^6 \left( \frac{6}{\sqrt{45}} \vec{i} + \frac{3}{\sqrt{45}} \vec{j} \right) \text{ N/C} E 1 = 45 ⋅ 1 0 − 6 81 ( 45 6 i + 45 3 j ) = 1.8 ⋅ 1 0 6 ( 45 6 i + 45 3 j ) N/C Vector desde q 2 q_2 q 2 a P:
r ⃗ 2 P = ( 0 − 6 ⋅ 10 − 3 ) i ⃗ + ( 3 ⋅ 10 − 3 − 0 ) j ⃗ = ( − 6 ⋅ 10 − 3 i ⃗ + 3 ⋅ 10 − 3 j ⃗ ) m \vec{r}_{2P} = (0 - 6 \cdot 10^{-3}) \vec{i} + (3 \cdot 10^{-3} - 0) \vec{j} = (-6 \cdot 10^{-3} \vec{i} + 3 \cdot 10^{-3} \vec{j}) \text{ m} r 2 P = ( 0 − 6 ⋅ 1 0 − 3 ) i + ( 3 ⋅ 1 0 − 3 − 0 ) j = ( − 6 ⋅ 1 0 − 3 i + 3 ⋅ 1 0 − 3 j ) m r ^ 2 P = − 6 ⋅ 10 − 3 i ⃗ + 3 ⋅ 10 − 3 j ⃗ 45 ⋅ 10 − 3 = − 6 45 i ⃗ + 3 45 j ⃗ \hat{r}_{2P} = \frac{-6 \cdot 10^{-3} \vec{i} + 3 \cdot 10^{-3} \vec{j}}{\sqrt{45} \cdot 10^{-3}} = -\frac{6}{\sqrt{45}} \vec{i} + \frac{3}{\sqrt{45}} \vec{j} r ^ 2 P = 45 ⋅ 1 0 − 3 − 6 ⋅ 1 0 − 3 i + 3 ⋅ 1 0 − 3 j = − 45 6 i + 45 3 j E ⃗ 2 = K q 2 r P 2 r ^ 2 P = ( 9 ⋅ 10 9 ) 9 ⋅ 10 − 9 ( 45 ⋅ 10 − 3 ) 2 ( − 6 45 i ⃗ + 3 45 j ⃗ ) N/C \vec{E}_2 = K \frac{q_2}{r_P^2} \hat{r}_{2P} = (9 \cdot 10^9) \frac{9 \cdot 10^{-9}}{(\sqrt{45} \cdot 10^{-3})^2} \left( -\frac{6}{\sqrt{45}} \vec{i} + \frac{3}{\sqrt{45}} \vec{j} \right) \text{ N/C} E 2 = K r P 2 q 2 r ^ 2 P = ( 9 ⋅ 1 0 9 ) ( 45 ⋅ 1 0 − 3 ) 2 9 ⋅ 1 0 − 9 ( − 45 6 i + 45 3 j ) N/C E ⃗ 2 = 1.8 ⋅ 10 6 ( − 6 45 i ⃗ + 3 45 j ⃗ ) N/C \vec{E}_2 = 1.8 \cdot 10^6 \left( -\frac{6}{\sqrt{45}} \vec{i} + \frac{3}{\sqrt{45}} \vec{j} \right) \text{ N/C} E 2 = 1.8 ⋅ 1 0 6 ( − 45 6 i + 45 3 j ) N/C El campo eléctrico total en el punto P es la suma vectorial de E ⃗ 1 \vec{E}_1 E 1 y E ⃗ 2 \vec{E}_2 E 2 :
E ⃗ P = E ⃗ 1 + E ⃗ 2 = 1.8 ⋅ 10 6 [ ( 6 45 i ⃗ + 3 45 j ⃗ ) + ( − 6 45 i ⃗ + 3 45 j ⃗ ) ] \vec{E}_P = \vec{E}_1 + \vec{E}_2 = 1.8 \cdot 10^6 \left[ \left( \frac{6}{\sqrt{45}} \vec{i} + \frac{3}{\sqrt{45}} \vec{j} \right) + \left( -\frac{6}{\sqrt{45}} \vec{i} + \frac{3}{\sqrt{45}} \vec{j} \right) \right] E P = E 1 + E 2 = 1.8 ⋅ 1 0 6 [ ( 45 6 i + 45 3 j ) + ( − 45 6 i + 45 3 j ) ] E ⃗ P = 1.8 ⋅ 10 6 [ ( 6 45 − 6 45 ) i ⃗ + ( 3 45 + 3 45 ) j ⃗ ] \vec{E}_P = 1.8 \cdot 10^6 \left[ \left( \frac{6}{\sqrt{45}} - \frac{6}{\sqrt{45}} \right) \vec{i} + \left( \frac{3}{\sqrt{45}} + \frac{3}{\sqrt{45}} \right) \vec{j} \right] E P = 1.8 ⋅ 1 0 6 [ ( 45 6 − 45 6 ) i + ( 45 3 + 45 3 ) j ] E ⃗ P = 1.8 ⋅ 10 6 [ 0 ⋅ i ⃗ + 6 45 j ⃗ ] = 1.8 ⋅ 10 6 ⋅ 6 45 j ⃗ N/C \vec{E}_P = 1.8 \cdot 10^6 \left[ 0 \cdot \vec{i} + \frac{6}{\sqrt{45}} \vec{j} \right] = 1.8 \cdot 10^6 \cdot \frac{6}{\sqrt{45}} \vec{j} \text{ N/C} E P = 1.8 ⋅ 1 0 6 [ 0 ⋅ i + 45 6 j ] = 1.8 ⋅ 1 0 6 ⋅ 45 6 j N/C E ⃗ P = 10.8 45 ⋅ 10 6 j ⃗ N/C \vec{E}_P = \frac{10.8}{\sqrt{45}} \cdot 10^6 \vec{j} \text{ N/C} E P = 45 10.8 ⋅ 1 0 6 j N/C E ⃗ P ≈ 1.61 ⋅ 10 6 j ⃗ N/C \vec{E}_P \approx 1.61 \cdot 10^6 \vec{j} \text{ N/C} E P ≈ 1.61 ⋅ 1 0 6 j N/C 
