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Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge. Find the magnitude of the electric field at the center of the square. (k=1/4πϵ0=8.99×109 N.m^2/C^2)

2 Answers

6 votes

Answer:

E = 440816.32 N/C

Step-by-step explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m


K =1/4 \pi \epsilon_0 = 8.99 * 10^9 N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have


E_1 + E_3 = 0


E =E_2 + E_4


E = 2 E_2

[
E_2 =(2* k * q)/(r^2)

[
r= ((0.5^2 + 0.5^2)^2)/(2) = 0.35 m]

plugging all value


E = 2 E_2


E = 2 E_2 =(2* k * q)/(r^2)


E = (2 * 8.99 * 10^93* 10^(-6))/(0.35^2)

E = 440816.32 N/C

User Hmahdavi
by
3.3k points
3 votes

Answer:


E=4.32* 10^(5)\ N.C^(-1)

Step-by-step explanation:

Given:

  • charges at the each of the three corner of a square,
    q_1=q_2=q_3=+3* 10^(-6)\ C
  • side of the square,
    a=0.5\ m
  • charge at the remaining corner of the square,
    q_4=-3* 10^(-6)\ C

Distance of the center of the square from each of the vertex of square:

Using Pythagoras theorem:


d^2+d^2=a^2


2d^2=0.5^2


d=0.3536\ m

As there two equal like charges at an equal distance on the opposite ends of a diagonal so they will cancel out the effect of field due to each other.

Electric field at the center of the square due to
q_4:


E_4=(1)/(4\pi.\epsilon_0) * (q_4)/(d^2)


E_4=9* 10^9*(3* 10^(-6))/(0.125)


E_4=216*10^(3)\ N.C^(-1)

The charge on the opposite vertex will have the equal effect in the same direction which is towards the charge
q_4. (refer the attached schematic)

So, the net electric field at the center:


E=2* E_4


E=2* 216* 10^3


E=4.32* 10^(5)\ N.C^(-1)

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The-example-1
User David Callanan
by
3.0k points