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Two point charges 1 = +2.0 and 2 = 3. 0 are separated by a

distance of 40 .
Determine the
() The resultant electric field strength at point X
() The electric force that acts on a point charge = 0.50 placed
at X

User Flyingzl
by
7.9k points

1 Answer

4 votes

To determine the resultant electric field strength at point X, we need to calculate the individual electric field strengths created by each point charge and then add them as vectors.

The electric field strength created by a point charge is given by the equation:

E = k * |q| / r^2

where E is the electric field strength, k is the electrostatic constant (k ≈ 8.99 × 10^9 N m^2/C^2), q is the magnitude of the charge, and r is the distance from the charge.

For charge 1 (q1 = +2.0) at point X, the distance is 40 units:

E1 = (8.99 × 10^9 N m^2/C^2) * |2.0 C| / (40 m)^2

E1 = 11243750 N/C

For charge 2 (q2 = -3.0) at point X, the distance is also 40 units:

E2 = (8.99 × 10^9 N m^2/C^2) * |3.0 C| / (40 m)^2

E2 = 16865625 N/C

To determine the resultant electric field at point X, we need to add the two electric field vectors:

EResultant = E1 + E2

EResultant = 11243750 N/C + 16865625 N/C

EResultant = 28109375 N/C

Therefore, the resultant electric field strength at point X is 28109375 N/C.

To determine the electric force on a point charge (q3 = 0.50) placed at point X, we can use the equation:

F = q3 * EResultant

F = (0.50 C) * (28109375 N/C)

F = 14054687.5 N

Therefore, the electric force acting on the point charge at point X is 14054687.5 Newtons.

User Mythofechelon
by
8.2k points

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