A uniformly charged thin disc of radius R has surface charge density σ. A point on its axis is at distance x from the centre. The value of x for which the axial electric field equals one-half of its value at the centre is:
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Question

A uniformly charged thin disc of radius R has surface charge density σ. A point on its axis is at distance x from the centre. The value of x for which the axial electric field equals one-half of its value at the centre is:

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G1 Physics · Accepted Answer

E(x)=σ/(2ε₀)[1-x/√(x²+R²)] and E(0)=σ/(2ε₀). Setting E(x)=E(0)/2 gives x/√(x²+R²)=1/2. Thus 4x²=x²+R², so x=R/√3.