Question

Two charges of opposite signs but unknown magnitudes are separated by a distance, r, along the x-axis. The positive charge, q​1​​, is to the left of the negative charge, q​2​​, as shown in the figure below. Consider three regions along the x-axis : (I) to the left of the positive charge, (II) in between the two charges, and (III) to the right of the negative charge. Considering just these two charges, in which of these regions could the electric field possibly be zero.

Region I Region II Region IIl 91 q2

Answer 1

Electric field is given by

E = k*Q/R^2

Now direction of electric field will be towards the negative charge and away from positive charge. So Using this information:

E1 = Electric field due to q1

E2 = Electric field due to q2

In region 1:

Direction of E1 = Since q1 is +ve, So Electric field will be away from charge in left side (in -ve x-axis)

Direction of E2 = Since q2 is -ve, So Electric field will be towards the charge in right side (in +ve x-axis)

So Net electric field will be

Enet = E2 - E1

Now from above equation when E2 = E1, net electric field could be zero.

In region 2:

Direction of E1 = Since q1 is +ve, So Electric field will be away from charge in right side (in +ve x-axis)

Direction of E2 = Since q2 is -ve, So Electric field will be towards the charge in right side (in +ve x-axis)

So Net electric field will be

Enet = E1 + E2

Since sum of two values cannot be zero, So in this region net electric field can not be zero.

In region 3:

Direction of E1 = Since q1 is +ve, So Electric field will be away from charge in right side (in +ve x-axis)

Direction of E2 = Since q2 is -ve, So Electric field will be towards the charge in left side (in -ve x-axis)

So Net electric field will be

Enet = E1 - E2

Now from above equation when E1 = E2, net electric field could be zero.

So In region 1 and 3 Electric field could be zero.

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