Question

Theory Electric field is a measure of the degree to which a source charge Q distorts its surrounding space. As a result of this distortion, the motion of other charges q are driven by the electric force Fe the source charge exerts on them. In this regard, electric field is defined as the ratio of the force (which the source charge Q exerts other charge q) to charge q: F Ē 9 The direction of electric field at a point in space is determined by the direction of force acting on a positive test charge brought to that point. The magnitude of electric field is determined by E = 1 Q 476 r2 (81. =6) 1 Experiments 1.1 Point Charge Plot the electric field vectors at several points around: 1. A positive point charge q= +4.0 x 10-6C at several distances from 3.0 cm to 12 cm. Use a scale of 1.0cm equaling 10' N/C or other scale. 2. A negative point charge q = -4.0 x 10-6C at several distances from 3.0 cm to 12 cm. Use a scale of 1.0 cm equaling 10' N/C or other scale. + . O . 3 cm 6 cm

Answer 1

+

The E field of a positive point charge points radially outwards as shown above. The magnitude of the E field is constant for every point of a given circunference centered at the point charge, and it falls with the square of the distance:

$E(r=3cm)=9\times 10^{9}\frac{Nm^{2}}{C^{2}}\times\frac{4\times 10^{-6}C}{(0.03m)^{2}}=4\times 10^{7}\frac{N}{C}$

$E(r=6cm)=9\times 10^{9}\frac{Nm^{2}}{C^{2}}\times\frac{4\times 10^{-6}C}{(0.06m)^{2}}=1\times 10^{7}\frac{N}{C}$

$E(r=9cm)=9\times 10^{9}\frac{Nm^{2}}{C^{2}}\times\frac{4\times 10^{-6}C}{(0.09m)^{2}}=4.4\times 10^{6}\frac{N}{C}$

E(r = 12cm) = 9 X 109-/m² N X 4 x 10-60 (0.12m) = 2.5 x 106

For a negative charge we get the same pattern, only now the direction of the E field is radially inwards