Question

A charge Q located at a point "r" is in equilibrium under the combined electric field of three charges qi, q2 and q3 . If the charges q1 and q2 are located at points r1 and r2 respectively, find the direction of the force Q due to q3in terms of q1, q2, q3,r, r1 andr2,

Answer 1

Given:  A charge Q located at a point "r",three charges q1, q2 and q3,  charges q1 and q2 are located at points r1 and r2

To find: find the direction of the force Q due to q3in terms of q1, q2 and r, r1 and r2.

Solution:

• To find the direction of forces, we first need to find the force.

• So,

• Let, F1 be the force on charge Q due to q1

• F2 be the force on charge Q due to q2

• F3 be the force on charge Q due to q3 .

• As the system is in equilibrium, so sum of all the forces will be zero.

                    F1 + F2 + F3 = 0

• Applying values of forces, we get:

              K x Q x q1 x (r(cap) - r1(cap)) / {mod(r(cap) - r1(cap))}³  

             +  K x Q x q2 x (r(cap) - r2(cap)) / {mod(r(cap) - r2(cap))}³  +  F3 = 0

• So,

             F3 = -KQ x [ q1 x (r(cap) - r1(cap)) / {mod(r(cap) - r1(cap))}³

              +  q2 x (r(cap) - r2(cap)) / {mod(r(cap) - r2(cap))}³ ]

Answer:

So, direction of force is

          -KQ x [ q1 x (r(cap) - r1(cap)) / {mod(r(cap) - r1(cap))}³

                     +  q2 x (r(cap) - r2(cap)) / {mod(r(cap) - r2(cap))}³ ]

Answer 2

The direction of the force Q due to q3 in terms of q1, q2, q3, $\overrightarrow{\mathrm{r}}$, $\overrightarrow{\mathrm{r1}}$ and $\overrightarrow{\mathrm{r2}}$ is given as:

$-k q\left[\left(\frac{q_{1}(\vec{r}-\overrightarrow{r_{1}})}{\left|\vec{r}-\vec{r}_{1}\right|^{3}}\right)+\left(\frac{q_{2}(\vec{r}-\overrightarrow{r_{2}})}{\left|\vec{r}-\vec{r}_{2}\right|^{3}}\right)\right]$

Explanation:

When the system is in equilibrium, the total force is given as:

$\overrightarrow{F_{1}}+\overrightarrow{F_{2}}+\overrightarrow{F_{3}}=0$

Where,

$\overrightarrow{F_{1}}$ = Force on charge Q due to q1

$\overrightarrow{F_{2}}$ = Force on charge Q due to q2

$\overrightarrow{F_{3}}$ = Force on charge Q due to q3

Now, the force on charge Q due to q1 is:

$\overrightarrow{F_{1}}=\frac{k q_{1} Q}{|\vec{r}-\overrightarrow{r_{1}}|^{2}} \times \frac{\vec{r}-\overrightarrow{r_{1}}}{|\vec{r}-\overrightarrow{r_{1}}|}$

$\therefore \vec{F}_{1}=\frac{k q_{1} Q\left(\vec{r}-\vec{r}_{1}\right)}{\left|\vec{r}-\vec{r}_{1}\right|^{3}}$

The force on charge Q due to q2 is:

$\overrightarrow{F_{2}}=\frac{k q_{2} Q}{|\vec{r}-\overrightarrow{r_{2}}|^{2}} \times \frac{\vec{r}-\overrightarrow{r_{2}}}{|\vec{r}-\overrightarrow{r_{2}}|}$

$\therefore \overrightarrow{F_{2}}=\frac{k q_{2} Q(\vec{r}-\overrightarrow{r_{2}})}{|\vec{r}-\overrightarrow{r_{2}}|^{3}}$

Now,

$\overrightarrow{F_{3}}=-(\overrightarrow{F_{1}}+\overrightarrow{F_{2}})$

Thus,

$\overrightarrow{F_{3}}=-\left[\left(\frac{k q_{1} Q(\vec{r}-\overrightarrow{r_{1}})}{|\vec{r}-\overrightarrow{r_{1}}|^{3}}\right)+\left(\frac{k q_{2} Q(\vec{r}-\overrightarrow{r_{2}})}{|\vec{r}-\overrightarrow{r_{2}}|^{3}}\right)\right]$

$\therefore \overrightarrow{F_{3}}=-k q\left[\left(\frac{q_{1}(\vec{r}-\overrightarrow{r_{1}})}{|\vec{r}-\overrightarrow{r_{1}}|^{3}}\right)+\left(\frac{q_{2}(\vec{r}-\overrightarrow{r_{2}})}{|\vec{r}-\overrightarrow{r_{2}}|^{3}}\right)\right]$