Question

two charges +5C and +15C are located at points (2,-4,3) and (-3,2,1).the force acting on +15C charge ......

Answer 1

Answer:

F =Qq/r2

F=5C×15C/(2-1)²

F=75/1

F=75N

Answer 2

The force acting on +15C is 1.288 × 10^9(5i - 6j + 2k).

Two charges +5C and +15C are located at points (2, -4, 3) and (-3, 2, 1).

We have to find the force acting on +15C charge.

Here, Q₁ = +5C and Q₂ = +15C are located at points, r₁ = (2, -4, 3) and r₂ = (-3, 2, 1).

∴ The force acting on +15C charge by +5C = F₂₁ =

$\frac{KQ_1Q_2}{|r_{12}|^3}(\vec{r_1}-\vec{r_2})$

$\vec{r_1}-\vec{r_2}=(2+3)\hat{i}+(-4-2)\hat{j}+(3-1)\hat{k}\\=5\hat{i}-6\hat{j}+2\hat{k})$

$|r_{12}|=\sqrt{5^2+(-6)^2+(2)^2}=\sqrt{65}$

$F_{21}=\frac{9\times10^9\times15\times5}{65\sqrt{65}}(5\hat{i}-6\hat{j}+2\hat{k}$

$=1.288\times10^9(5\hat{i}-6\hat{j}+\hat{k}$

Therefore the force acting on +15C is 1.288 × 10^9(5i - 6j + 2k).

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