1.6 Four point charges ${\mathit{q}}_{\mathit{A}}$ = 2 μC, ${\mathit{q}}_{\mathit{B}}$ = –5 μC,  ${\mathit{q}}_{\mathit{C}}$= 2 μC, and ${\mathit{q}}_{\mathit{D}}$ = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

1.6

 

In the adjoining figure ABCD is a square with sides AB = BC = CD = DA = 10 cm

Diagonals, AC = BD  = $\sqrt[]{{10}^2+{10}^2}$ = $10\sqrt[]{2}$cm

AO = OC = DO = BO = $5\sqrt[]{2}$ cm

At the centre of the square ABCD, O, a charge of 1 $\mu Cisplaced.$$\text{exception 25:}$

The force of repulsion between the charges placed at A and at O is equal in magnitude but opposite in direction between the charges placed at point C and centre O. Similarly ,the force of attraction between the charges placed at B & O and D & O will be equal in magnitude but opposite in direction. These charges will cancel each other.

Hence, the net charge at centre O will be zero.

Here is a deeper explanation. 

What you just saw is the us

...more

1.6

 

In the adjoining figure ABCD is a square with sides AB = BC = CD = DA = 10 cm

Diagonals, AC = BD  = $\sqrt[]{{10}^2+{10}^2}$ = $10\sqrt[]{2}$cm

AO = OC = DO = BO = $5\sqrt[]{2}$ cm

At the centre of the square ABCD, O, a charge of 1 $\mu Cisplaced.$$\text{exception 25:}$

The force of repulsion between the charges placed at A and at O is equal in magnitude but opposite in direction between the charges placed at point C and centre O. Similarly ,the force of attraction between the charges placed at B & O and D & O will be equal in magnitude but opposite in direction. These charges will cancel each other.

Hence, the net charge at centre O will be zero.

Here is a deeper explanation. 

What you just saw is the use of the superposition principle. The total force on the 1 μC charge at the centre (O) is the vector sum of the forces from each of the four corner charges. 

Pairs can break down the analysis. 

• Forces from qA and qC: The force exerted by qA (+2 μC) on the central charge is repulsive. The force from qC (+2 μC) is also repulsive. Now, since qA and qC are equal and are at an equal distance from the centre, the forces they exert are equal in magnitude. But they point in opposite directions along the diagonal AC. So, they cancel each other out.

• Forces from qB and qD: Similarly, the force exerted by qB (–5 μC) is attractive, and the force from qD (–5 μC) is also attractive. Because these charges are equal and equidistant from the centre, the forces are equal in magnitude but opposite in direction along the diagonal BD. That causes them to cancel out as well.

As both pairs of forces nullify each other, the net electrostatic force on the central charge is zero. Read more about Coulomb's Law

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<p><strong>1.6</strong></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728885558phpgkhKVI_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728885558phpgkhKVI.jpeg" alt="" width="1200" height="675"></picture></div></div><p>&nbsp;</p><p>In the adjoining figure ABCD is a square with sides AB = BC = CD = DA = 10 cm</p><p>Diagonals, AC = BD &nbsp;=&nbsp;<span title="Click to copy mathml"><math><mroot><mrow><mrow><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow></mrow></mroot></math></span> = <span title="Click to copy mathml"><math><mn>10</mn><mroot><mrow><mrow><mn>2</mn></mrow></mrow><mrow></mrow></mroot></math></span>cm</p><p>AO = OC = DO = BO = <span contenteditable="false"> <math> <mn>5</mn> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span>cm</p><p>At the centre of the square ABCD, O, a charge of 1 <span title="Click to copy mathml"><math><mi>μ</mi><mi>C</mi><mi></mi><mi>i</mi><mi>s</mi><mi></mi><mi>p</mi><mi>l</mi><mi>a</mi><mi>c</mi><mi>e</mi><mi>d</mi><mo>.</mo></math></span><math><mroot><mrow></mrow></mroot></math></p><p>The force of repulsion between the charges placed at A and at O is equal in magnitude but opposite in direction between the charges placed at point C and centre O. Similarly ,the force of attraction between the charges placed at B &amp; O and D &amp; O will be equal in magnitude but opposite in direction. These charges will cancel each other.</p><p>Hence, the net charge at centre O will be zero.<br><br>Here is a deeper explanation.&nbsp;</p><p dir="ltr">What you just saw is the use of the superposition principle. The total force on the 1 μC charge at the centre (O) is the vector sum of the forces from each of the four corner charges.&nbsp;</p><p dir="ltr">Pairs can break down the analysis.&nbsp;</p><ul><li dir="ltr" aria-level="1"><p dir="ltr" role="presentation">Forces from qA and qC: The force exerted by qA (+2 μC) on the central charge is repulsive. The force from qC (+2 μC) is also repulsive. Now, since qA and qC are equal and are at an equal distance from the centre, the forces they exert are equal in magnitude. But they point in opposite directions along the diagonal AC. So, they cancel each other out.</p></li><li dir="ltr" aria-level="1"><p dir="ltr" role="presentation">Forces from qB and qD: Similarly, the force exerted by qB (–5 μC) is attractive, and the force from qD (–5 μC) is also attractive. Because these charges are equal and equidistant from the centre, the forces are equal in magnitude but opposite in direction along the diagonal BD. That causes them to cancel out as well.</p></li></ul><p dir="ltr">As both pairs of forces nullify each other, the net electrostatic force on the central charge is zero. Read more about &lt;a href="https://www.shiksha.com/preparation/physics-electric-charge-and-field-coulombs-law-5676-tp" target="_blank" rel="noopener"&gt;Coulomb's Law&lt;/a&gt;</p>

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