SATHEE: Related Problems with Solution

Related Problems with Solution

Problem 1 : Two point charges, Q₁ and Q₂, are placed 2 meters apart. Q₁ is +4 μC and Q₂ is -3 μC. Calculate the electric field at a point on the line joining the two charges and 1 meter from Q₁.

Solution :

To find the electric field at the given point, we’ll calculate the electric field due to each charge individually and then combine them using vector addition.

Electric field due to Q₁ at the given point: $[E_{1} = \frac{k \cdot | Q_{1} |}{r^{2}} = \frac{8.99 \times 10^{9} \cdot 4 \times 10^{- 6}}{(1)^{2}} = 35.96, N/C] (d i r e c t e d t o t h e r i g h t)$

Electric field due to Q₂ at the given point: $[E_{2} = \frac{k \cdot | Q_{2} |}{r^{2}} = \frac{8.99 \times 10^{9} \cdot 3 \times 10^{- 6}}{(1)^{2}} = 26.97, N/C] (d i r e c t e d t o t h e l e f t)$

Now, find the net electric field: $[E_{net} = E_{1} + E_{2} = 35.96, N/C - 26.97, N/C = 8.99, N/C]$

So, the electric field at the given point is 8.99 N/C to the right.

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Related Problems with Solution

Problem 1 : Two point charges, Q1 and Q2, are placed 2 meters apart. Q1 is +4 μC and Q2 is -3 μC. Calculate the electric field at a point on the line joining the two charges and 1 meter from Q1.

Solution :

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Problem 1 : Two point charges, Q₁ and Q₂, are placed 2 meters apart. Q₁ is +4 μC and Q₂ is -3 μC. Calculate the electric field at a point on the line joining the two charges and 1 meter from Q₁.